Derivatives of Trigonometric Functions

Chapter Summary (Express)

Greek letters being usual to denote angles, we will take as the usual letter for any variable angle the letter $θ$ (“theta”). In this chapter, $θ$ is measured in radians.¹

Derivative of Sine

We are investigating the derivative of the function

y = \sin θ

\frac{d (\sin θ)}{d θ}

. The task involves exploring the relationship between small changes in the angle

θ

(denoted

d θ

) and the corresponding small changes in its sine value (denoted

d y

)

d y = \sin (θ + d θ) - \underset{y}{\underset{⏟}{\sin θ}} .

Using the difference to product identity, we derive

d y = 2 \cos (θ + \frac{1}{2} d θ) \cdot \sin \frac{d θ}{2} .

Finally, considering

d θ

as indefinitely small and using the approximation

\sin \frac{d θ}{2} \approx \frac{d θ}{2}

, we establish

\frac{d y}{d θ} = \cos θ .

Let us consider the function $y = \sin θ .$

What we have to investigate is the value of $\frac{d (\sin θ)}{d θ}$ ; or, in other words, if the angle $θ$ varies, we have to find the relation between the increment of the sine and the increment of the angle, both increments being indefinitely small in themselves. Examine Fig. 15.1, wherein, if the radius of the circle is unity, the height of $y$ is the sine, and $θ$ is the angle. Now, if $θ$ is supposed to increase by the addition to it of the small angle $d θ$ —an element of angle—the height of $y$ , the sine, will be increased by a small element $d y$ . The new height $y + d y$ will be the sine of the new angle $θ + d θ$ , or, stating it as an equation, $y + d y = \sin (θ + d θ);$ and subtracting from this the first equation gives $d y = \sin (θ + d θ) - \sin θ .$

Unit circle diagram with radius 1 showing angle θ from the positive x-axis. The vertical height (y) represents sin θ. Shows a small angular increment dθ and the corresponding increment dy in the sine value, illustrating the relationship between the angle change and the change in sine value. — Fig. 15.1

The quantity on the right-hand side is the difference between two sines, and books on trigonometry tell us how to work this out. For they tell us that if $M$ and $N$ are two different angles, $\sin M - \sin N = 2 \cos \frac{M + N}{2} \cdot \sin \frac{M - N}{2} .$

If, then, we put $M = θ + d θ$ for one angle, and $N = θ$ for the other, we may write \begin{align} dy &= 2 \cos\frac{\theta + d\theta + \theta}{2} \cdot \sin\frac{\theta + d\theta - \theta}{2}, \end{align} or, \begin{align} dy &= 2\cos\left(\theta + \frac{1}{2}d\theta\right)\cdot \sin\frac{d\theta}{2}. \end{align}

But if we regard $d θ$ as indefinitely small, then in the limit we may neglect $\frac{1}{2} d θ$ by comparison with $θ$ , and may also take $\sin \frac{d θ}{2}$ as being the same as $\frac{1}{2} d θ$ . The equation then becomes: \begin{align} dy &= 2 \cos \theta \times \frac{1}{2} d \theta; \\ dy &= \cos \theta \cdot d \theta, \end{align} and, finally, \begin{align} \dfrac{dy}{d \theta} &= \cos \theta. \end{align} [Notice that the approximation $\sin \frac{d θ}{2} \approx \frac{d θ}{2}$ is true only when $d θ$ is measured in radians.]

The accompanying curves, Fig. 15.2 and Fig. 15.3, show, plotted to scale, the values of $y = \sin θ$ , and $\frac{d y}{d θ} = \cos θ$ , for the corresponding values of $θ$ .

Graph of y = sin θ plotted to scale, showing the sinusoidal wave pattern that oscillates between -1 and 1 with period 2π radians. The curve starts at (0,0), reaches maximum at (π/2,1), crosses zero at (π,0)$, reaches minimum at (3π/2,-1), and completes one cycle at (2π,0). — Fig. 15.2

Graph of dy/dθ = cos θ plotted to scale, showing the cosine wave pattern that oscillates between -1 and 1 with period 2π radians. The curve starts at (0,1), crosses zero at (π/2,0), reaches minimum at (π,-1), crosses zero again at (3π/2,0), and returns to maximum at (2π,1). This represents the derivative of the sine function shown in Fig. 15.2. — Fig. 15.3

Derivative of Cosine

To differentiate

y = \cos θ

, since

\cos θ = \sin (\frac{π}{2} - θ)

, we can use the Chain Rule to get

\frac{d y}{d θ} = - \cos (\frac{π}{2} - θ) = - \sin θ .

Take next the cosine.

Let $y = \cos θ$ .

Now $\cos θ = \sin (\frac{π}{2} - θ)$ .

Therefore \begin{align} &\begin{align} dy = d\left(\sin\left(\frac{\pi}{2} - \theta\right)\right) &= \cos\left(\frac{\pi}{2} - \theta\right) \times d(-\theta), \\ &= \cos\left(\frac{\pi}{2} - \theta\right) \times (-d\theta), \end{align} \\ &\frac{dy}{d\theta} = -\cos\left(\frac{\pi}{2} - \theta\right). \end{align} And it follows that $\frac{d y}{d θ} = - \sin θ .$

Derivative of Tangent

\tan θ = \frac{\sin θ}{\cos θ}

, we can apply the Quotient Rule and the identity

\sin^{2} θ + \cos^{2} θ = 1

to find

\frac{d (\tan θ)}{d θ}

\frac{d (\tan θ)}{d θ} = \sec^{2} θ = 1 + \tan^{2} θ .

Lastly, take the tangent.

As $\tan θ = \frac{\sin θ}{\cos θ}$ , we can apply the Quotient Rule to find $\frac{d (\tan θ)}{d θ}$ :²

\begin{align} \frac{d(\tan\theta)}{d\theta}&=\frac{\cos\theta \dfrac{d(\sin\theta)}{d\theta}-\sin\theta \dfrac{d(\cos\theta)}{d\theta}}{\cos^2\theta}\\ &=\frac{\cos\theta \cdot \cos\theta -\sin\theta (-\sin\theta)}{\cos^2\theta}\\ &=\frac{\cos^2\theta+\sin^2\theta}{\cos^2\theta} \end{align} Since $\frac{\cos^{2} θ + \sin^{2} θ}{\cos^{2} θ} = 1 + {(\frac{\sin θ}{\cos θ})}^{2}$ we get $\frac{d (\tan θ)}{d θ} = 1 + \tan^{2} θ .$ Also, since $\sin^{2} θ + \cos^{2} θ = 1$ , we get $\frac{d (\tan θ)}{d θ} = \frac{1}{\cos^{2} θ} = \sec^{2} θ .$ Therefore, $\frac{d (\tan θ)}{d θ} = \sec^{2} θ = 1 + \tan^{2} θ .$

Summary of Results

Collecting these results, we have: $\begin{array}{cc} y & \frac{d y}{d θ} \\ \sin θ & \cos θ \\ \cos θ & - \sin θ \\ \tan θ & \sec^{2} θ = 1 + \tan^{2} θ \end{array}$

The results tabulated above are true only when

θ

is measured in radians.

To derive the above results, we replaced $\sin (d θ / 2)$ by $d θ / 2$ . In general, $\sin x$ is approximately equal to $x$ when (1) $x$ is small (2) $x$ is measured in radians $\sin x \approx x (x is small and measured in radians) .$ For example, $1^{\circ}$ is the same as $\frac{π}{180}$ radians, and we cannot approximate $\sin 1^{\circ}$ by 1 $\sin 1^{\circ} \neq 1$ but $\sin 1^{\circ} = \sin \frac{π}{180} \approx \frac{π}{180} = 0.017 .$ Therefore, the results tabulated above are true only when $θ$ is measured in radians.

Sometimes, in mechanical and physical questions, as, for example, in simple harmonic motion and in wave-motions, we have to deal with angles that increase in proportion to the time. Thus, if $T$ be the time of one complete period, or movement round the circle, then, since the angle all round the circle is $2 π$ radians, (equivalent to $360^{\circ}$ ), the amount of angle moved through in time $t$ , will be \begin{align} \theta &= 2\pi\frac{t}{T},\quad \text{in radians.} \end{align} If the frequency, or number of periods per second, be denoted by $n$ , then $n = \frac{1}{T}$ , and we may then write: $θ = 2 π n t .$ Then we shall have $y = \sin (2 π n t) .$

If, now, we wish to know how the sine varies with respect to time, we must differentiate with respect, not to $θ$ , but to $t$ . For this we must resort to the Chain Rule, and put $\frac{d y}{d t} = \frac{d y}{d θ} \cdot \frac{d θ}{d t} .$

Now $\frac{d θ}{d t}$ will obviously be $2 π n$ ; so that \begin{align} \frac{dy}{dt} &= \cos \theta \times 2\pi n \\ &= 2\pi n \cdot \cos (2\pi nt). \end{align} Similarly, it follows that \begin{align} \frac{d\left(\cos (2\pi nt)\right)}{dt} &= -2\pi n \cdot \sin (2\pi nt). \end{align}

Second Derivative of Sine or Cosine.

\frac{d^{2} (\sin θ)}{d θ^{2}} = \frac{d (\cos θ)}{d t} = - \sin θ .

and

\frac{d^{2} (\cos θ)}{d θ^{2}} = \frac{d (- \sin θ)}{d t} = - \cos θ .

We have seen that when $\sin θ$ is differentiated with respect to $θ$ it becomes $\cos θ$ ; and that when $\cos θ$ is differentiated with respect to $θ$ it becomes $- \sin θ$ ; or, in symbols, $\frac{d^{2} (\sin θ)}{d θ^{2}} = - \sin θ .$

So we have this curious result that we have found a function such that if we differentiate it twice over, we get the same thing from which we started, but with the sign changed from $+$ to $-$ .

The same thing is true for the cosine; for differentiating $\cos θ$ gives us $- \sin θ$ , and differentiating $- \sin θ$ gives us $- \cos θ$ ; or thus: $\frac{d^{2} (\cos θ)}{d θ^{2}} = - \cos θ .$

Sines and cosines are the only functions of which the second derivative is equal (and of opposite sign to) the original function.

Examples

With what we have so far learned we can now differentiate expressions of a more complex nature.

Example 15.1. If $y = \arcsin x$ , find $\frac{d y}{d x}$ .

[In many modern calculus textbooks, the inverse sine is denoted by $\sin^{- 1}$ ; that is, $\arcsin x = \sin^{- 1} x$ . Note that $\sin^{- 1} x$ is NOT the same as $\frac{1}{\sin x}$ . To avoid confusion, the notation $\arcsin x$ may be preferred over $\sin^{- 1} x$ in some texts, including this one.]

Solution

If $y$ is the arc whose sine is $x$ , then $x = \sin y$ . $\frac{d x}{d y} = \cos y .$

Passing now from the inverse function to the original one, we get \begin{align} \frac{dy}{dx} &= \frac{1}{\;\dfrac{dx}{dy}\;} = \frac{1}{\cos y}. \end{align} now since $\cos^{2} y + \sin^{2} y = 1$ , $\cos y = \pm \sqrt{1 - \sin^{2} y} = \pm \sqrt{1 - x^{2}};$ hence \begin{align} \frac{dy}{dx} &=\frac{1}{\cos y}= \pm \frac{1}{\sqrt{1-x^2}}. \end{align} But which one is correct? plus or minus? If we look at the graph of $y = \arcsin x$ (see the following figure), we realize that the slope of the curve is always positive, indicating that that we must take the positive square root. Hence, $\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}} .$

Graph of y = arcsin x showing a continuous, increasing curve defined on the domain [-1,1] with range [−π/2, π/2]. The curve passes through the origin (0, 0), with endpoints at (−1,−π/2) and (1,π/2). The slope is positive throughout the domain, confirming that the derivative should take the positive square root. — Fig. 15.4

\begin{align} \boxed{\dfrac{d(\arcsin x)}{dx}=\dfrac{1}{\sqrt{1-x^2}}} \end{align}

Example 15.2. If $y = \cos^{3} θ$ , find $\frac{d y}{d θ}$ .

Solution

This is the same thing as $y = (\cos θ)^{3}$ .

Let $\cos θ = v$ ; then $y = v^{3}$ ; $\frac{d y}{d v} = 3 v^{2}$ . \begin{align} \frac{dv}{d\theta} &= -\sin\theta.\\ \frac{dy}{d\theta} &= \frac{dy}{dv} \times \frac{dv}{d\theta}\\ &= -3 \cos^2 \theta \sin\theta. \end{align}

Example 15.3. If $y = \sin (x + a)$ , find $\frac{d y}{d x}$ .

Solution

Let $x + a = v$ ;then $y = \sin v$ . $\frac{d y}{d v} = \cos v; \frac{d v}{d x} = 1 and \frac{d y}{d x} = \cos (x + a) .$

Example 15.4. If $y = \ln \sin θ$ , find $\frac{d y}{d θ}$ .

Solution

Let $\sin θ = v$ ; $y = \ln v$ . \begin{align} \frac{dy}{dv} &= \frac{1}{v};\quad \frac{dv}{d\theta}=\cos\theta;\\ \frac{dy}{d\theta} &= \frac{1}{\sin\theta} \times \cos\theta = \cot\theta. \end{align}

Example 15.5. If $y = \cot θ = \frac{\cos θ}{\sin θ}$ , find $\frac{d y}{d θ}$ .

Solution

\begin{align} \frac{dy}{d\theta} &= \frac{-\sin^2\theta - \cos^2 \theta}{\sin^2 \theta}\\ &= -(1+\cot^2 \theta) = -\text{csc}^2 \theta. \end{align}

Example 15.6. If $y = \tan 3 θ$ , find $\frac{d y}{d θ}$ .

Solution

Let $3 θ = v$ ; $y = \tan v$ ; $\frac{d y}{d v} = \sec^{2} v$ . $\frac{d v}{d θ} = 3; \frac{d y}{d θ} = 3 \sec^{2} 3 θ .$

Example 15.7. If $y = \sqrt{1 + 3 \tan^{2} θ}$ , find $\frac{d y}{d θ}$ .

Solution

$y = (1 + 3 \tan^{2} θ)^{\frac{1}{2}}$ .

Let $3 \tan^{2} θ = v$ . $y = (1 + v)^{\frac{1}{2}}; \frac{d y}{d v} = \frac{1}{2 \sqrt{1 + v}};$ $\frac{d v}{d θ} = 6 \tan θ \sec^{2} θ$ (for, if $\tan θ = u$ , \begin{align} v &= 3u^2;\quad \frac{dv}{du} = 6u;\quad \frac{du}{d\theta} = \sec^2 \theta; \end{align} hence $\frac{d v}{d θ} = 6 \tan θ \sec^{2} θ$ ;)
hence \begin{align} \frac{dy}{d\theta} &= \frac{6\tan\theta \sec^2\theta}{2\sqrt{1 + 3\tan^2\theta}}. \end{align}

Example 15.8. If $y = \sin x \cos x$ , find $\frac{d y}{d x}$ .

Solution

\begin{align} \frac{dy}{dx} &= \sin x(-\sin x) + \cos x \times \cos x = \cos^2 x - \sin^2 x. \end{align}

Exercises

Exercise 15.1. Differentiate the following: \begin{align} \text{(i)}\quad y &= A \sin\left(\theta - \frac{\pi}{2}\right).\\ \text{(ii)}\quad y &= \sin^2 \theta;\quad \text{and }\quad y = \sin 2\theta.\\ \text{(iii)}\quad y &= \sin^3 \theta;\quad \text{and }\quad y = \sin 3\theta. \end{align}

Answer

(i) $\frac{d y}{d θ} = A \cos (θ - \frac{π}{2})$ ;

(ii) $\frac{d y}{d θ} = 2 \sin θ \cos θ = \sin 2 θ$ and $\frac{d y}{d θ} = 2 \cos 2 θ$ ;

(iii) $\frac{d y}{d θ} = 3 \sin^{2} θ \cos θ$ and $\frac{d y}{d θ} = 3 \cos 3 θ$ .

Solution

(i) $y = A \sin (θ - \frac{π}{2})$

We write $y = A \sin u where u = θ - \frac{π}{2}$ Using the Chain Rule: \begin{align} \frac{d y}{d \theta}&= \frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ & =(A \cos u)(1) \\ & =A \cos \left(\theta-\frac{\pi}{2}\right) \end{align}

(ii) If $y = \sin^{2} θ = (\sin θ)^{2}$

Let $y = u^{2} where u = \sin θ$ Using the Chain Rule: \begin{align} \frac{d y}{d \theta} & =\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ & =2 u \cdot \cos \theta \\ & =2 \sin \theta \cos \theta \end{align} The result can also be written as $\sin 2 θ$ since $\sin 2 θ = 2 \sin θ \cos θ .$

If $y = \sin 2 θ$ , let $y = \sin u where u = 2 θ .$ Using the Chain Rule: \begin{align} \frac{d y}{d \theta}&=\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ &=(\cos u)(2) \\ &=2 \cos (2 \theta) \end{align} If $y = \sin^{3} θ = (\sin θ)^{3}$ , we write $y = u^{3} where u = \sin θ$ Then using the Chain Rule \begin{align} \frac{d y}{d \theta}&=\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ &=3 u^{2} \cdot \cos \theta \\ &=3(\sin \theta)^{2} \cdot \cos \theta \\ &=3 \cos \theta \sin ^{2} \theta \\ &=(\cos u) \cdot(3) \\ &=3 \cos 3 \theta \cdot \sin 3 \theta \end{align} If $y = \sin 3 θ$ , we write $y = \sin u where u = 3 θ$ Then \begin{align} \frac{dy}{d \theta}&=\frac{d y}{d u} \cdot \frac{d u}{d \theta} \\ &=(\cos u)(3)\\ &=3\cos {3\theta} \end{align}

Exercise 15.2. Find the value of $θ$ for which $\sin θ \times \cos θ$ is a maximum.

Answer

$θ = 45^{\circ}$ or $\frac{π}{4}$ radians.

Solution

$y = \sin θ \cos θ$

Method 1) Using the Product Rule:

\begin{align} \frac{d y}{d \theta}&=\cos \theta \cdot \cos \theta-\sin \theta \sin \theta \\ &=\cos ^{2} \theta-\sin ^{2} \theta \\ &=\cos 2 \theta \end{align} $\frac{d y}{d θ} = \cos 2 θ = 0 \Leftrightarrow 2 θ = \frac{π}{2} or 2 θ = \frac{3 π}{2}$ $\frac{d y}{d θ} = 0 \Leftrightarrow θ = \frac{π}{4} or θ = \frac{3 π}{4}$ \begin{align} \frac{d^{2} y}{d \theta^{2}}&= \frac{d(\cos 2 \theta)}{d(2 \theta)} \cdot \frac{d(2 \theta)}{d \theta} \\ &=(-\sin 2 \theta)(2) \\ &=-2 \sin 2 \theta \end{align} When $θ = \frac{π}{4}$ $\frac{d^{2} y}{d θ^{2}} = - 2 < 0$ Hence, the curve is concave downward, and thus when $θ = \frac{π}{4}$ , $y$ has a maximum of $\sin (\frac{π}{4}) \cos (\frac{π}{4}) = \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2} .$

When $θ = \frac{3 π}{4}$ $\frac{d^{2} y}{d θ^{2}} = - 2 \sin (\frac{3 π}{2}) = 2 > 0$ hence, the curve is concave upward and thus when $θ = \frac{3 π}{4}$ , $y$ has a minimum of \begin{align} \sin \left(\frac{3 \pi}{4}\right) \cos \left(\frac{3 \pi}{4}\right) & =\sin \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \cos \left(\frac{\pi}{2}+\frac{\pi}{4}\right) \\ & =\cos \left(\frac{\pi}{4}\right) \times\left(-\sin \left(\frac{\pi}{4}\right)\right) \\ & =\frac{1}{\sqrt{2}} \times \frac{-1}{\sqrt{2}} \\ & =-\frac{1}{2}. \end{align}

Method 2)

$y = \sin θ \cdot \cos θ = \frac{1}{2} \sin 2 θ$ $y$ is a maximum wherever $\sin 2 θ$ is a maximum and that occurs when $2 θ = \frac{π}{2} or θ = \frac{π}{4}$ The maximum of $y$ is then $\frac{1}{2} \cdot \sin (\frac{π}{2}) = \frac{1}{2}$ .

Exercise 15.3. Differentiate $y = \frac{1}{2 π} \cos (2 π n t)$ .

Answer

$\frac{d y}{d t} = - n \sin 2 π n t$ .

Solution

$y = \frac{1}{2 π} \cos (2 π n t)$ We write $y = \frac{1}{2 π} \cos u where u = 2 π n t$ \begin{align} \frac{d y}{d t} &= \frac{d y}{d u} \cdot \frac{d u}{d t} \\ &=-\frac{1}{2 \pi} \sin u \cdot 2 \pi n \\ &=-n \sin u \\ &= n \sin (2 \pi n t) \end{align}

Exercise 15.4. If $y = \sin a^{x}$ , find $\frac{d y}{d x}$ .

Answer

$a^{x} \ln a \cos a^{x}$ .

Solution

$y = \sin (a^{x})$ Let $y = \sin u$ where $u = a^{x}$ . Then \begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\cos u \cdot a^{x} \cdot \ln a \\ & =\cos \left(a^{x}\right) \cdot a^{x} \cdot \ln a \end{align}

Exercise 15.5. Differentiate $y = \ln \cos x$ .

Answer

$\frac{\cos x}{\sin x} = \cot x$

Solution

$y = \ln \cos x$

We write $y = \ln u$ where $u = \cos x$ . Then

\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\frac{1}{u} \cdot(-\sin x) \\ & =\frac{1}{\cos x}(-\sin x) \\ & =-\tan x \end{align}

Exercise 15.6. Differentiate $y = 18.2 \sin (x + 26)$ .

Answer

$18.2 \cos (x + 26)$ .

Solution

$y = 18.2 \sin (x + 26)$

We wrtie $y = 18.2 \sin u$ where $u = x + 26$ . Then

\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =(18.2 \cos u)(1) \\ & =18.2 \cos (x+26) \end{align}

Exercise 15.7. Plot the curve $y = 10 \sin (θ - \frac{π}{12})$ ; and show that the slope of the curve at $θ = \frac{5 π}{12}$ is half the maximum slope.

Answer

The slope is $\frac{d y}{d θ} = 100 \cos (θ - \frac{π}{12})$ , which is a maximum when $(θ - \frac{π}{12}) = 0$ , or $θ = \frac{π}{12}$ ; the value of the slope being then $= 100$ . When $θ = \frac{5 π}{12}$ the slope is $100 \cos (\frac{5 π}{12} - \frac{π}{12}) = 100 \cos \frac{π}{3} = 100 \times \frac{1}{2} = 50$ .

Solution

\begin{align} & y=10 \sin \left(\theta-\frac{\pi}{12}\right) \\ & \frac{d y}{d \theta}=10 \cos \left(\theta-\frac{\pi}{12}\right) \end{align}

To find the maximum slope, we have to differentiate $\frac{d y}{d θ}$ with respect to $θ$ and equate the result to zero

$\frac{d (\frac{d y}{d θ})}{d θ} = \frac{d^{2} y}{d θ^{2}} = - 10 \sin (θ - \frac{π}{12}) = 0$

\begin{align} \frac{d^{2} y}{d \theta^{2}}=0\quad\Leftrightarrow &\quad \theta-\frac{\pi}{12}=0 \quad \text { or } \quad \theta-\frac{\pi}{12}=\pi \\ \frac{d^{2} y}{d \theta^{2}}=0\quad \Leftrightarrow &\quad \theta=\frac{\pi}{12} \quad \text { or } \quad \theta=\frac{13 \pi}{12} \end{align}

When $θ = \frac{π}{12}$ $\frac{d y}{d θ} = 10 \cos 0 = 10 (max slope)$

When $θ = \frac{13 π}{12}$ $\frac{d y}{d θ} = 10 \cos (π) = - 10 (min slope)$

Slope when $θ = \frac{5 π}{12}$ : \begin{align} \frac{d y}{d \theta} & =10 \cos \left(\frac{5 \pi}{12}-\frac{\pi}{12}\right)=10 \cos \left(\frac{\pi}{3}\right) \\ & =10 \times \frac{1}{2}=5 . \end{align} As we can see the slope of the curve at $θ = \frac{5 π}{12}$ , which is 5, is half the maximum slope, which is 10 and occurs when $θ = \frac{π}{12} .$

Exercise 15.8. If $y = \sin θ \cdot \sin 2 θ$ , find $\frac{d y}{d θ}$ .

Answer

\begin{align} \cos\theta \sin2\theta + 2\cos2\theta \sin\theta &= 2\sin\theta\left(\cos^2 \theta + \cos2\theta\right) \\ &= 2\sin\theta\left(3\cos^2 \theta - 1\right). \end{align}

Solution

$y = \sin θ \cdot \sin 2 θ$

Using the Product Rule:

$\frac{d y}{d θ} = \frac{d (\sin θ)}{d θ} \cdot \sin 2 θ + \sin θ \frac{d (\sin 2 θ)}{d θ}$

We showed in Exercise 1 (ii) $\frac{d (\sin 2 θ)}{d θ} = 2 \cos 2 θ$ .

Hence

$\frac{d y}{d θ} = \cos θ \cdot \sin 2 θ + 2 \sin θ \cos 2 θ$

We can further simplify it using

$\sin 2 θ = 2 \sin θ \cos θ$ and $\cos 2 θ = 2 \cos^{2} θ - 1.$

\begin{align} \frac{d y}{d \theta} & =2 \sin \theta \cos ^{2} \theta+2 \sin \theta\left(2 \cos ^{2} \theta-1\right) \\ & =2 \sin \theta\left(\cos ^{2} \theta+2 \cos ^{2} \theta-1\right) \\ & =2 \sin \theta\left(3 \cos ^{2} \theta-1\right) . \end{align}

Exercise 15.9. If $y = a \cdot \tan^{m} (θ^{n})$ , find the derivative of $y$ with respect to $θ$ .

Answer

$a m n θ^{n - 1} \tan^{m - 1} (θ^{n}) \sec^{2} θ^{n}$ .

Solution

$y = a \tan^{m} (θ^{n}) = a {[\tan (θ^{n})]}^{m}$ We write $y = a u^{m} where u = \tan v and v = θ^{n} .$ Then

\begin{align} & \frac{d y}{d \theta}=\frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d \theta} \\ &=a m u^{m-1} \cdot \sec ^{2} v \cdot n \cdot \theta^{n-1} \\ &=a m (\tan v)^{m-1} \cdot \sec ^{2} v \cdot n \cdot \theta^{n-1} \\ &=a \cdot m \cdot n \cdot\left[\tan \left(\theta^{n}\right)\right]^{m-1} \cdot \sec ^{2}\left(\theta^{n}\right) \cdot \theta^{n-1} \end{align} Note that $\sec^{2} (θ^{n})$ means ${[\sec (θ^{n})]}^{2}$ and ${[\tan (θ^{n})]}^{m - 1}$ can be written as $\tan^{m - 1} (θ^{n})$ . Hence $\frac{d y}{d θ} = a m n θ^{n - 1} \tan^{m - 1} (θ^{n}) \sec^{2} (θ^{n})$

Exercise 15.10. If $y = e^{x} \sin^{2} x$ , find $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ .

Answer

$e^{x} (\sin^{2} x + \sin 2 x)$ ; $e^{x} (\sin^{2} x + 2 \sin 2 x + 2 \cos 2 x)$ .

Solution

$y = e^{x} \sin^{2} x = e^{x} (\sin x)^{2}$

$\frac{d y}{d x} = \frac{d (e^{x})}{d x} \sin^{2} x + e^{x} \frac{d (\sin^{2} x)}{d x}$

To find $\frac{d (\sin^{2} x)}{d x}$ , we notice that

$\sin^{2} x = (\sin x)^{2}$

and

\begin{align} \frac{d\left(u^{2}\right)}{d x} & =\frac{d\left(u^{2}\right)}{d u} \cdot \frac{d u}{d x} \qquad(u=\sin x) \\ & =2 u \cdot \cos x \\ & =2 \sin x \cos x \\ & =\sin 2 x \end{align} Therefore \begin{align} \frac{d y}{d x} & =\frac{d\left(e^{x}\right)}{d x} \sin ^{2} x+e^{x} \frac{d\left(\sin ^{2} x\right)}{d x} \\ & =e^{x} \sin ^{2} x+e^{x} \sin 2 x \\ & =e^{x}\left(\sin ^{2} x+\sin 2 x\right) \end{align}

The second derivative: \begin{align} \frac{d y}{d x}&= \frac{d\left(e^{x}\left(\sin ^{2} x+\sin 2 x\right)\right)}{dx}\\ &=\frac{d(e^x)}{dx}\left(\sin ^{2} x+\sin 2 x\right)+e^x\left(\frac{d(\sin^2 x)}{dx}+\frac{d(\sin 2x)}{dx}\right)\\ &=e^x\left(\sin ^{2} x+\sin 2 x\right)+e^x\left(\underbrace{2\cos x \sin x}_{\sin 2x}+2\cos 2x\right)\\ &=e^x\left(\sin^2 x+2\sin 2x+2\cos 2x\right). \end{align}

Exercise 15.11. Differentiate the three equations of Exercises 14.II (see here), No. 4, and compare their derivatives, as to whether they are equal, or nearly equal, for very small values of $x$ , or for very large values of $x$ , or for values of $x$ in the neighbourhood of $x = b$ .

Answer

$(i) \frac{d y}{d x} = \frac{a b}{{(x + b)}^{2}}$ ; (ii) $\frac{a}{b} e^{- \frac{x}{b}}$ ; (iii) $\frac{2 a b}{π} \cdot \frac{1}{(b^{2} + x^{2})}$ .

Solution

(i) Using the Quotient Rule: $\frac{d y}{d x} = \frac{a (x + b) - a x}{(x + b)^{2}} = \frac{a b}{(x + b)^{2}}$

(ii) $\frac{d y}{d x} = - a \times (- \frac{1}{b}) e^{- \frac{x}{b}} = \frac{a}{b} e^{- \frac{x}{b}}$

(iii) To differentiate $y = \frac{2 a}{π} \arctan (\frac{x}{b})$ , we write $y = \frac{2 a}{π} \arctan u where u = \frac{x}{b} .$ Then \begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\frac{2 a}{\pi} \frac{1}{1+u^{2}} \cdot \frac{1}{b} \\ & =\frac{2 a}{\pi b} \frac{1}{1+\left(\frac{x}{b}\right)^{2}} \\ & =\frac{2 a b}{\pi} \frac{1}{x^{2}+b^{2}} \end{align}

When $x$ is very large

\begin{align} & \frac{a b}{(x+b)^{2}} \approx \frac{a b}{x^{2}} \approx 0 \\ & \frac{a}{b} e^{-\frac{x}{b}} \approx \frac{a}{b} \times 0=0 \\ & \frac{2 a b}{x^{2}+b^{2}} \approx \frac{2 a b}{x^{2}} \approx 0 \end{align}

Therefore their slopes are almost zero for large values of $x$ .

When $x \approx 0$ \begin{align} & \frac{a b}{(x+b)^{2}} \approx \frac{a b}{b^{2}}=\frac{a}{b} \\ & e^{-\frac{x}{b}} \approx 1 \Rightarrow \frac{a}{b} e^{-\frac{x}{b}} \approx \frac{a}{b} \\ & \frac{2 a b}{x^{2}+b^{2}} \approx \frac{2 a b}{b^{2}}=\frac{2 a}{b} \end{align} When $x$ is small ( $x \approx 0$ ), the slopes of $y = \frac{a x}{x + b}$ and $y = a (1 - e^{- \frac{x}{b}})$ are almost identical, but the slope of $\frac{2 a}{π} \arctan (\frac{x}{b})$ is twice as much as them.

When $x \approx b$ \begin{align} & \frac{a b}{(x+b)^{2}} \approx \frac{a b}{(2 b)^{2}}=\frac{a}{4 b} \\ & \frac{a}{b} e^{-\frac{x}{b}} \approx \frac{a}{b} \cdot e^{-1}=\frac{a}{2.78 b} \\ & \frac{2 a b}{x^{2}+b^{2}} \approx \frac{2 a b}{2 b^{2}}=\frac{a}{b} . \end{align}

Exercise 15.12. Differentiate the following: \begin{align} \text{(i)}\quad y &= \sec x. & \text{(ii)}\quad y &= \arccos x. \\ \text{(iii)}\quad y &= \arctan x. & \text{(iv)}\quad y &= \text{arcsec} x. \\ \text{(v)}\quad y &= \tan x \times \sqrt{3 \sec x}. && \end{align}

Answer

(i) $\frac{d y}{d x} = \sec x \tan x$ ;

(ii) $\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^{2}}}$ ;

(iii) $\frac{d y}{d x} = \frac{1}{1 + x^{2}}$ ;

(iv) $\frac{d y}{d x} = \frac{1}{| x | \sqrt{x^{2} - 1}}$ ;

(v) $\frac{d y}{d x} = \frac{\sqrt{3 \sec x} (3 \sec^{2} x - 1)}{2}$ .

Solution

(i) $y = \sec x = \frac{1}{\cos x}$
Using the Quotient Rule \begin{align} \frac{d y}{d x} & =\frac{-(-\sin x)}{\cos ^{2} x}=\frac{\sin x}{\cos x} \cdot \frac{1}{\cos x} \\ & =\tan x \cdot \sec x . \end{align}

(ii) $y = \arccos x$ (or $y = \cos^{- 1} x$ )

If $y = \arccos x$ , then $x = \cos y$ and $\frac{d x}{d y} = - \sin y$

\begin{align} \frac{d y}{d x} & =\frac{1}{\frac{d x}{d y}} \\ & =\frac{1}{-\sin y} \end{align} Since $\sin y = \pm \sqrt{1 - \cos^{2} y}$ $\frac{d y}{d x} = \frac{1}{- \sqrt{1 - \cos^{2} y}}$ Since $x = \cos y$ $\frac{d y}{d x} = \frac{1}{\mp \sqrt{1 - x^{2}}}$ But which one is correct? The $-$ sign or the $+$ sign? If we look at the graph of $y = \arccos x$ , the slope is negative everywhere.

Hence $\frac{d (\arccos x)}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}}$

(iii) If $y = \arctan x$ , then $x = \tan y$ and

$\frac{d x}{d y} = 1 + \tan^{2} y (or \sec^{2} y)$ Therefore,

\begin{align} \frac{d y}{d x} & =\frac{1}{\dfrac{d x}{d y}} \\ & =\frac{1}{1+\tan ^{2} y} \\ & =\frac{1}{1+x^{2}} \end{align}

(iv) If $y = arcsec x$ then $x = \sec y$ . In part (i), we showed that

$\frac{d x}{d y} = \tan y \cdot \sec y .$ Hence \begin{align} \frac{d y}{d x} & =\frac{1}{\dfrac{d x}{d y}} \\ & =\frac{1}{\tan y \cdot \sec y} \end{align} Since $1 + \tan^{2} y = \sec^{2} y \Rightarrow \tan y = \pm \sqrt{\sec^{2} y - 1}$ , we have \begin{align} \frac{d y}{d x} & =\frac{1}{ \pm \sqrt{\sec ^{2} y-1} \cdot \sec y} \\ & =\frac{1}{ \pm x \sqrt{x^{2}-1}} . \end{align}

Now we need to decide about the sign.

As we can see from the graph $y = arcsec x$ , the slope is always positive. So we must have

\begin{align} & \frac{d y}{d x}=\frac{1}{x \sqrt{x^{2}-1}} &&\text { if } x \geq 1 \\ & \frac{d y}{d x}=\frac{1}{-x \sqrt{x^{2}-1}} &&\text { if } x \leq-1 \end{align} We can combine these two and write $\frac{d y}{d x} = \frac{1}{| x | \sqrt{x^{2} - 1}}$

(v) $y = \tan x \times \sqrt{3 \sec x} = \sqrt{3} \tan x \cdot \sqrt{\sec x}$ .

Using the Product Rule: $\frac{d y}{d x} = \sqrt{3} [\frac{d (\tan x)}{d x} \cdot \sqrt{\sec x} + \tan x \frac{d (\sqrt{\sec x})}{d x}]$

To find $\frac{d (\sqrt{\sec x})}{d x}$ , let $u = \sec x$ . Then

\begin{align} \frac{d(\sqrt{u})}{d x} & =\frac{d(\sqrt{u})}{d u} \cdot \frac{d u}{d x} \\ & =\frac{1}{2 \sqrt{u}} \cdot \sec x \cdot \tan x \\ & =\frac{1}{2 \sqrt{\sec x}} \cdot \sec x \cdot \tan x \\ & =\frac{1}{2} \tan x\cdot \sqrt{\sec x} . \end{align}

Hence

\begin{align} \frac{d y}{d x} & =\sqrt{3}\left[\sec ^{2} x \sqrt{\sec x}+\frac{1}{2} \tan ^{2} x \sqrt{\sec x}\right] \\ & =\sqrt{3 \sec x} \quad\left[\sec ^{2} x+\frac{1}{2} \tan ^{2} x\right] \end{align}

We can simplify this further (or rewrite it in a different form) using $1 + \tan^{2} x = \sec^{2} x$

\begin{align} & \frac{d y}{d x}=\sqrt{3 \sec x} \cdot\left[\sec ^{2} x+\frac{1}{2}\left(\sec ^{2} x-1\right)\right] \\ & =\frac{1}{2} \sqrt{3 \sec x}\left[3 \sec ^{2} x-1\right] \end{align}

Exercise 15.13. Differentiate $y = \sin (2 θ + 3)^{2.3}$ .

Answer

$\frac{d y}{d θ} = 4.6 {(2 θ + 3)}^{1.3} \cos {(2 θ + 3)}^{2.3}$ .

Solution

\begin{align} & y=\sin (2 \theta+3)^{2.3}=\sin \left[(2 \theta+3)^{2.3}\right] \\ y & =\sin u, \quad u=v^{2.3}, \quad v=2 \theta+3 \\ \frac{d y}{d x}= & \frac{d y}{d u} \cdot \frac{d u}{d v} \cdot \frac{d v}{d x} \\ & =\cos u\left(2.3 v^{1.3}\right) \cdot 2 \\ & =4.6(2 \theta+3)^{1.3} \cos v^{2.3} \\ & =4.6(2 \theta+3)^{1.3} \cos \left[(2 \theta+3)^{2.3}\right] . \end{align}

Exercise 15.14. Differentiate $y = θ^{3} + 3 \sin (θ + 3) - 3^{\sin θ} - 3^{θ}$ .

Answer

$\frac{d y}{d θ} = 3 θ^{2} + 3 \cos (θ + 3) - \ln 3 (\cos θ \times 3^{\sin θ} + 3 θ)$ .

Solution

$y = θ^{3} + 3 \sin (θ + 3) - 3^{\sin θ} - 3^{θ}$

We differentiate term by term

$\frac{d y}{d θ} = 3 θ^{2} + 3 \cos (θ + 3) - \frac{d (3^{\sin θ})}{d θ} - 3^{θ} \cdot \ln 3$

To find $\frac{d (3^{\sin θ})}{d θ}$ , let $\sin θ = u$ and

\begin{align} \frac{d 3^{u}}{d \theta} & =\frac{d 3^{u}}{d u} \cdot \frac{d u}{d \theta} \\ & =3^{u} \cdot \ln 3 \cdot \cos \theta \\ & =3^{\sin \theta} \cdot \ln 3 \cdot \cos \theta \end{align}

Therefore

$\frac{d y}{d θ} = 3 θ^{2} + 3 \cos (θ + 3) - \ln 3 \cos θ 3^{\sin θ} - \ln 3 \cdot 3^{θ}$

Exercise 15.15. Find the maximum or minimum of $y = θ \cos θ$ ( $- \frac{π}{2} < θ < \frac{π}{2}$ ).

Answer

$θ = \cot θ; θ = \pm 0.86$ ; $y = \pm 0.56$ is max. for $+ θ$ , min. for $- θ$ .

Solution

$y = θ \cos θ$

Using the Product Rule

\begin{align} \frac{d y}{d \theta}&=\cos \theta+\theta(-\sin \theta) \\ &=\cos \theta-\theta \sin \theta \\ \end{align}

$\frac{d y}{d θ} = \cos θ - θ \sin θ = 0 \Rightarrow θ = \cot θ$

The solutions of $θ = \cot θ$ can be obtained approximately

$θ \approx \pm 0.86$

To distinguish between a maximum and a minimum, we use the Second Derivative Test:

\begin{align} \frac{d^{2} y}{d \theta^{2}} & =\frac{d(\cos \theta-\theta \sin \theta)}{d \theta} \\ & =-\sin \theta-(\sin \theta+\theta \cos \theta) \\ & =\theta \cos \theta-2 \sin \theta \end{align}

When $θ \approx 0.86$

$\frac{d^{2} y}{d θ^{2}} \approx - 0.95 < 0$ Hence, the curve is concave downward, and $y$ has a maximum of $0.86 \cos (0.86) \approx 0.56$ when $θ \approx 0.86$ .

When $θ \approx - 0.86$ $\frac{d^{2} y}{d θ^{2}} \approx 0.95 > 0$ Hence, the curve is concave upward, and thus $y$ has a minimum of $- 0.86 \cos (- 0.86) \approx - 0.56$ when $θ \approx - 0.86 .$

In higher mathematics, angles are always measured in radians, unless a different unit of measurement is explicitly specified.↩︎

Alternatively, let \begin{align} y &= \tan \theta, \\ dy &= \tan(\theta + d\theta) - \tan\theta. \end{align}Expanding, as shown in books on trigonometry, \begin{align} \tan(\theta + d\theta) &= \frac{\tan\theta + \tan d\theta} {1 - \tan\theta\cdot\tan d\theta}; \end{align} [Recall that $\tan (A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ ]
whence \begin{align} dy &= \frac{\tan\theta + \tan d\theta} {1-\tan\theta\cdot\tan d\theta} - \tan\theta \\ &= \frac{(1 + \tan^2\theta)\tan d\theta} {1-\tan\theta\cdot\tan d\theta}. \end{align}

Now remember that if $d θ$ is indefinitely diminished, the value of $\tan d θ$ becomes identical with $d θ$ , and $\tan θ \cdot d θ$ is negligibly small compared with $1$ , so that the expression reduces to \begin{align} dy &= \frac{(1+\tan^2 \theta)\, d\theta}{1}, \end{align} so that $\frac{d y}{d θ} = 1 + \tan^{2} θ,$ or $\frac{d y}{d θ} = \sec^{2} θ .$ ↩︎

Full Chapter

Greek letters being usual to denote angles, we will take as the usual letter for any variable angle the letter $θ$ (“theta”). In this chapter, $θ$ is measured in radians.¹

Derivative of Sine

Let us consider the function $y = \sin θ .$

What we have to investigate is the value of $\frac{d (\sin θ)}{d θ}$ ; or, in other words, if the angle $θ$ varies, we have to find the relation between the increment of the sine and the increment of the angle, both increments being indefinitely small in themselves. Examine the next figure, wherein, if the radius of the circle is unity, the height of $y$ is the sine, and $θ$ is the angle. Now, if $θ$ is supposed to increase by the addition to it of the small angle $d θ$ —an element of angle—the height of $y$ , the sine, will be increased by a small element $d y$ . The new height $y + d y$ will be the sine of the new angle $θ + d θ$ , or, stating it as an equation, $y + d y = \sin (θ + d θ);$ and subtracting from this the first equation gives $d y = \sin (θ + d θ) - \sin θ .$

The accompanying curves in the next two figures show, plotted to scale, the values of $y = \sin θ$ , and $\frac{d y}{d θ} = \cos θ$ , for the corresponding values of $θ$ .

Derivative of Cosine

Take next the cosine.

Let $y = \cos θ$ .

Now $\cos θ = \sin (\frac{π}{2} - θ)$ .

Therefore \begin{align} dy = d\left(\sin\left(\frac{\pi}{2} - \theta\right)\right) &= \cos\left(\frac{\pi}{2} - \theta\right) \times d(-\theta), \\ &= \cos\left(\frac{\pi}{2} - \theta\right) \times (-d\theta), \end{align} $\frac{d y}{d θ} = - \cos (\frac{π}{2} - θ) .$ And it follows that $\frac{d y}{d θ} = - \sin θ .$

Derivative of Tangent

Lastly, take the tangent.

As $\tan θ = \frac{\sin θ}{\cos θ}$ , we can apply the Quotient Rule to find $\frac{d (\tan θ)}{d θ}$ :²

Summary of Results

Collecting these results, we have: $\begin{array}{cc} y & \frac{d y}{d θ} \\ \sin θ & \cos θ \\ \cos θ & - \sin θ \\ \tan θ & \sec^{2} θ = 1 + \tan^{2} θ \end{array}$

If, now, we wish to know how the sine varies with respect to time, we must differentiate with respect, not to $θ$ , but to $t$ . For this we must resort to the chain rule explained in the chapter on the Chain Rule, and put $\frac{d y}{d t} = \frac{d y}{d θ} \cdot \frac{d θ}{d t} .$

Second Derivatives of Sine and Cosine

So we have this curious result that we have found a function such that if we differentiate it twice over, we get the same thing from which we started, but with the sign changed from $+$ to $-$ .

Sines and cosines are the only functions of which the second derivative is equal (and of opposite sign to) the original function.

Examples

With what we have so far learned we can now differentiate expressions of a more complex nature.

Example 15.1. If $y = \arcsin x$ , find $\frac{d y}{d x}$ .

Solution. If $y$ is the arc whose sine is $x$ , then $x = \sin y$ . $\frac{d x}{d y} = \cos y .$

Passing now from the inverse function to the original one, we get \begin{align} \frac{dy}{dx} &= \frac{1}{\;\dfrac{dx}{dy}\;} = \frac{1}{\cos y}. \end{align} now since $\cos^{2} y + \sin^{2} y = 1$ , $\cos y = \pm \sqrt{1 - \sin^{2} y} = \pm \sqrt{1 - x^{2}};$ hence \begin{align} \frac{dy}{dx} &=\frac{1}{\cos y}= \pm \frac{1}{\sqrt{1-x^2}}. \end{align} But which one is correct? plus or minus? If we look at the graph of $y = \arcsin x$ (the following figure), we realize that the slope of the curve is always positive, indicating that that we must take the positive square root. Hence, $\frac{d y}{d x} = \frac{1}{\sqrt{1 - x^{2}}} .$

\begin{align} \boxed{\dfrac{d(\arcsin x)}{dx}=\dfrac{1}{\sqrt{1-x^2}}} \end{align}

Example 15.2. If $y = \cos^{3} θ$ , find $\frac{d y}{d θ}$ .

Solution. This is the same thing as $y = (\cos θ)^{3}$ .

Let $\cos θ = v$ ;then $y = v^{3}$ ; $\frac{d y}{d v} = 3 v^{2}$ . \begin{align} \frac{dv}{d\theta} &= -\sin\theta.\\ \frac{dy}{d\theta} &= \frac{dy}{dv} \times \frac{dv}{d\theta} = -3 \cos^2 \theta \sin\theta. \end{align}

Example 15.3. If $y = \sin (x + a)$ , find $\frac{d y}{d x}$ .

Solution. Let $x + a = v$ ;then $y = \sin v$ . $\frac{d y}{d v} = \cos v; \frac{d v}{d x} = 1 and \frac{d y}{d x} = \cos (x + a) .$

Example 15.4. If $y = \ln \sin θ$ , find $\frac{d y}{d θ}$ .

Solution. Let $\sin θ = v$ ; $y = \ln v$ . \begin{align} \frac{dy}{dv} &= \frac{1}{v};\quad \frac{dv}{d\theta}=\cos\theta;\\ \frac{dy}{d\theta} &= \frac{1}{\sin\theta} \times \cos\theta = \cot\theta. \end{align}

Example 15.5. If $y = \cot θ = \frac{\cos θ}{\sin θ}$ , find $\frac{d y}{d θ}$ .

Solution. \begin{align} \frac{dy}{d\theta} &= \frac{-\sin^2\theta - \cos^2 \theta}{\sin^2 \theta}\\ &= -(1+\cot^2 \theta) = -\text{csc}^2 \theta. \end{align}

Example 15.6. If $y = \tan 3 θ$ , find $\frac{d y}{d θ}$ .

Solution. Let $3 θ = v$ ; $y = \tan v$ ; $\frac{d y}{d v} = \sec^{2} v$ . $\frac{d v}{d θ} = 3; \frac{d y}{d θ} = 3 \sec^{2} 3 θ .$

Example 15.7. If $y = \sqrt{1 + 3 \tan^{2} θ}$ , find $\frac{d y}{d θ}$ .

Solution. $y = (1 + 3 \tan^{2} θ)^{\frac{1}{2}}$ .

Let $3 \tan^{2} θ = v$ . $y = (1 + v)^{\frac{1}{2}}; \frac{d y}{d v} = \frac{1}{2 \sqrt{1 + v}}$ $\frac{d v}{d θ} = 6 \tan θ \sec^{2} θ$ (for, if $\tan θ = u$ , \begin{align} v &= 3u^2;\quad \frac{dv}{du} = 6u;\quad \frac{du}{d\theta} = \sec^2 \theta; \end{align} hence $\frac{d v}{d θ} = 6 \tan θ \sec^{2} θ$ ;)
hence \begin{align} \frac{dy}{d\theta} &= \frac{6\tan\theta \sec^2\theta}{2\sqrt{1 + 3\tan^2\theta}}. \end{align}

Example 15.8. If $y = \sin x \cos x$ , find $\frac{d y}{d x}$ .

Solution. \begin{align} \frac{dy}{dx} &= \sin x(-\sin x) + \cos x \times \cos x = \cos^2 x - \sin^2 x. \end{align}

Exercises

Answer

(i) $\frac{d y}{d θ} = A \cos (θ - \frac{π}{2})$ ;

(ii) $\frac{d y}{d θ} = 2 \sin θ \cos θ = \sin 2 θ$ and $\frac{d y}{d θ} = 2 \cos 2 θ$ ;

(iii) $\frac{d y}{d θ} = 3 \sin^{2} θ \cos θ$ and $\frac{d y}{d θ} = 3 \cos 3 θ$ .

Solution

(i) $y = A \sin (θ - \frac{π}{2})$

(ii) If $y = \sin^{2} θ = (\sin θ)^{2}$

Exercise 15.2. Find the value of $θ$ for which $\sin θ \times \cos θ$ is a maximum.

Answer

$θ = 45^{\circ}$ or $\frac{π}{4}$ radians.

Solution

$y = \sin θ \cos θ$

Method 1) Using the Product Rule:

Method 2)

Exercise 15.3. Differentiate $y = \frac{1}{2 π} \cos (2 π n t)$ .

Answer

$\frac{d y}{d t} = - n \sin 2 π n t$ .

Solution

Exercise 15.4. If $y = \sin a^{x}$ , find $\frac{d y}{d x}$ .

Answer

$a^{x} \ln a \cos a^{x}$ .

Solution

Exercise 15.5. Differentiate $y = \ln \cos x$ .

Answer

$\frac{\cos x}{\sin x} = \cot x$

Solution

$y = \ln \cos x$

We write $y = \ln u$ where $u = \cos x$ . Then

\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =\frac{1}{u} \cdot(-\sin x) \\ & =\frac{1}{\cos x}(-\sin x) \\ & =-\tan x \end{align}

Exercise 15.6. Differentiate $y = 18.2 \sin (x + 26)$ .

Answer

$18.2 \cos (x + 26)$ .

Solution

$y = 18.2 \sin (x + 26)$

We wrtie $y = 18.2 \sin u$ where $u = x + 26$ . Then

\begin{align} \frac{d y}{d x} & =\frac{d y}{d u} \cdot \frac{d u}{d x} \\ & =(18.2 \cos u)(1) \\ & =18.2 \cos (x+26) \end{align}

Exercise 15.7. Plot the curve $y = 10 \sin (θ - \frac{π}{12})$ ; and show that the slope of the curve at $θ = \frac{5 π}{12}$ is half the maximum slope.

Answer

Solution

\begin{align} & y=10 \sin \left(\theta-\frac{\pi}{12}\right) \\ & \frac{d y}{d \theta}=10 \cos \left(\theta-\frac{\pi}{12}\right) \end{align}

To find the maximum slope, we have to differentiate $\frac{d y}{d θ}$ with respect to $θ$ and equate the result to zero

$\frac{d (\frac{d y}{d θ})}{d θ} = \frac{d^{2} y}{d θ^{2}} = - 10 \sin (θ - \frac{π}{12}) = 0$

When $θ = \frac{π}{12}$ $\frac{d y}{d θ} = 10 \cos 0 = 10 (max slope)$

When $θ = \frac{13 π}{12}$ $\frac{d y}{d θ} = 10 \cos (π) = - 10 (min slope)$

Illustration for Derivatives of Trigonometric Functions

Exercise 15.8. If $y = \sin θ \cdot \sin 2 θ$ , find $\frac{d y}{d θ}$ .

Answer

\begin{align} \cos\theta \sin2\theta + 2\cos2\theta \sin\theta &= 2\sin\theta\left(\cos^2 \theta + \cos2\theta\right) \\ &= 2\sin\theta\left(3\cos^2 \theta - 1\right). \end{align}

Solution

$y = \sin θ \cdot \sin 2 θ$

Using the Product Rule:

$\frac{d y}{d θ} = \frac{d (\sin θ)}{d θ} \cdot \sin 2 θ + \sin θ \frac{d (\sin 2 θ)}{d θ}$

We showed in Exercise 1 (ii) $\frac{d (\sin 2 θ)}{d θ} = 2 \cos 2 θ$ .

Hence

$\frac{d y}{d θ} = \cos θ \cdot \sin 2 θ + 2 \sin θ \cos 2 θ$

We can further simplify it using

$\sin 2 θ = 2 \sin θ \cos θ$ and $\cos 2 θ = 2 \cos^{2} θ - 1.$

Exercise 15.9. If $y = a \cdot \tan^{m} (θ^{n})$ , find the derivative of $y$ with respect to $θ$ .

Answer

$a m n θ^{n - 1} \tan^{m - 1} (θ^{n}) \sec^{2} θ^{n}$ .

Solution

$y = a \tan^{m} (θ^{n}) = a {[\tan (θ^{n})]}^{m}$ We write $y = a u^{m} where u = \tan v and v = θ^{n} .$ Then

Exercise 15.10. If $y = e^{x} \sin^{2} x$ , find $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ .

Answer

$e^{x} (\sin^{2} x + \sin 2 x)$ ; $e^{x} (\sin^{2} x + 2 \sin 2 x + 2 \cos 2 x)$ .

Solution

$y = e^{x} \sin^{2} x = e^{x} (\sin x)^{2}$

$\frac{d y}{d x} = \frac{d (e^{x})}{d x} \sin^{2} x + e^{x} \frac{d (\sin^{2} x)}{d x}$

To find $\frac{d (\sin^{2} x)}{d x}$ , we notice that

$\sin^{2} x = (\sin x)^{2}$

and

Answer

$(i) \frac{d y}{d x} = \frac{a b}{{(x + b)}^{2}}$ ; (ii) $\frac{a}{b} e^{- \frac{x}{b}}$ ; (iii) $\frac{2 a b}{π} \cdot \frac{1}{(b^{2} + x^{2})}$ .

Solution

(i) Using the Quotient Rule: $\frac{d y}{d x} = \frac{a (x + b) - a x}{(x + b)^{2}} = \frac{a b}{(x + b)^{2}}$

(ii) $\frac{d y}{d x} = - a \times (- \frac{1}{b}) e^{- \frac{x}{b}} = \frac{a}{b} e^{- \frac{x}{b}}$

When $x$ is very large

Therefore their slopes are almost zero for large values of $x$ .

Answer

(i) $\frac{d y}{d x} = \sec x \tan x$ ;

(ii) $\frac{d y}{d x} = - \frac{1}{\sqrt{1 - x^{2}}}$ ;

(iii) $\frac{d y}{d x} = \frac{1}{1 + x^{2}}$ ;

(iv) $\frac{d y}{d x} = \frac{1}{| x | \sqrt{x^{2} - 1}}$ ;

(v) $\frac{d y}{d x} = \frac{\sqrt{3 \sec x} (3 \sec^{2} x - 1)}{2}$ .

Solution

(ii) $y = \arccos x$ (or $y = \cos^{- 1} x$ )

If $y = \arccos x$ , then $x = \cos y$ and $\frac{d x}{d y} = - \sin y$

Hence $\frac{d (\arccos x)}{d x} = \frac{- 1}{\sqrt{1 - x^{2}}}$

(iii) If $y = \arctan x$ , then $x = \tan y$ and

$\frac{d x}{d y} = 1 + \tan^{2} y (or \sec^{2} y)$ Therefore,

\begin{align} \frac{d y}{d x} & =\frac{1}{\dfrac{d x}{d y}} \\ & =\frac{1}{1+\tan ^{2} y} \\ & =\frac{1}{1+x^{2}} \end{align}

(iv) If $y = arcsec x$ then $x = \sec y$ . In part (i), we showed that

Now we need to decide about the sign.

As we can see from the graph $y = arcsec x$ , the slope is always positive. So we must have

(v) $y = \tan x \times \sqrt{3 \sec x} = \sqrt{3} \tan x \cdot \sqrt{\sec x}$ .

Using the Product Rule: $\frac{d y}{d x} = \sqrt{3} [\frac{d (\tan x)}{d x} \cdot \sqrt{\sec x} + \tan x \frac{d (\sqrt{\sec x})}{d x}]$

To find $\frac{d (\sqrt{\sec x})}{d x}$ , let $u = \sec x$ . Then

Hence

We can simplify this further (or rewrite it in a different form) using $1 + \tan^{2} x = \sec^{2} x$

\begin{align} & \frac{d y}{d x}=\sqrt{3 \sec x} \cdot\left[\sec ^{2} x+\frac{1}{2}\left(\sec ^{2} x-1\right)\right] \\ & =\frac{1}{2} \sqrt{3 \sec x}\left[3 \sec ^{2} x-1\right] \end{align}

Exercise 15.13. Differentiate $y = \sin (2 θ + 3)^{2.3}$ .

Answer

$\frac{d y}{d θ} = 4.6 {(2 θ + 3)}^{1.3} \cos {(2 θ + 3)}^{2.3}$ .

Solution

Exercise 15.14. Differentiate $y = θ^{3} + 3 \sin (θ + 3) - 3^{\sin θ} - 3^{θ}$ .

Answer

$\frac{d y}{d θ} = 3 θ^{2} + 3 \cos (θ + 3) - \ln 3 (\cos θ \times 3^{\sin θ} + 3 θ)$ .

Solution

$y = θ^{3} + 3 \sin (θ + 3) - 3^{\sin θ} - 3^{θ}$

We differentiate term by term

$\frac{d y}{d θ} = 3 θ^{2} + 3 \cos (θ + 3) - \frac{d (3^{\sin θ})}{d θ} - 3^{θ} \cdot \ln 3$

To find $\frac{d (3^{\sin θ})}{d θ}$ , let $\sin θ = u$ and

\begin{align} \frac{d 3^{u}}{d \theta} & =\frac{d 3^{u}}{d u} \cdot \frac{d u}{d \theta} \\ & =3^{u} \cdot \ln 3 \cdot \cos \theta \\ & =3^{\sin \theta} \cdot \ln 3 \cdot \cos \theta \end{align}

Therefore

$\frac{d y}{d θ} = 3 θ^{2} + 3 \cos (θ + 3) - \ln 3 \cos θ 3^{\sin θ} - \ln 3 \cdot 3^{θ}$

Exercise 15.15. Find the maximum or minimum of $y = θ \cos θ$ ( $- \frac{π}{2} < θ < \frac{π}{2}$ ).

Answer

$θ = \cot θ; θ = \pm 0.86$ ; $y = \pm 0.56$ is max. for $+ θ$ , min. for $- θ$ .

Solution

$y = θ \cos θ$

Using the Product Rule

\begin{align} \frac{d y}{d \theta}&=\cos \theta+\theta(-\sin \theta) \\ &=\cos \theta-\theta \sin \theta \\ \end{align}

$\frac{d y}{d θ} = \cos θ - θ \sin θ = 0 \Rightarrow θ = \cot θ$

The solutions of $θ = \cot θ$ can be obtained approximately

$θ \approx \pm 0.86$

To distinguish between a maximum and a minimum, we use the Second Derivative Test:

\begin{align} \frac{d^{2} y}{d \theta^{2}} & =\frac{d(\cos \theta-\theta \sin \theta)}{d \theta} \\ & =-\sin \theta-(\sin \theta+\theta \cos \theta) \\ & =\theta \cos \theta-2 \sin \theta \end{align}

When $θ \approx 0.86$

$\frac{d^{2} y}{d θ^{2}} \approx - 0.95 < 0$ Hence, the curve is concave downward, and $y$ has a maximum of $0.86 \cos (0.86) \approx 0.56$ when $θ \approx 0.86$ .

When $θ \approx - 0.86$ $\frac{d^{2} y}{d θ^{2}} \approx 0.95 > 0$ Hence, the curve is concave upward, and thus $y$ has a minimum of $- 0.86 \cos (- 0.86) \approx - 0.56$ when $θ \approx - 0.86 .$

In higher mathematics, angles are always measured in radians, unless a different unit of measurement is explicitly specified.↩︎

Alternatively, let \begin{align} y &= \tan \theta, \\ dy &= \tan(\theta + d\theta) - \tan\theta. \end{align}Expanding, as shown in books on trigonometry, \begin{align} \tan(\theta + d\theta) &= \frac{\tan\theta + \tan d\theta} {1 - \tan\theta\cdot\tan d\theta}; \tag{Recall $\tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B}$} \end{align} whence \begin{align} dy &= \frac{\tan\theta + \tan d\theta} {1-\tan\theta\cdot\tan d\theta} - \tan\theta \\ &= \frac{(1 + \tan^2\theta)\tan d\theta} {1-\tan\theta\cdot\tan d\theta}. \end{align}

Chapter Summary (Express)

Derivative of Sine

Derivative of Cosine

Derivative of Tangent

Summary of Results

Second Derivative of Sine or Cosine.

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Derivative of Sine

Derivative of Cosine

Derivative of Tangent

Summary of Results

Second Derivatives of Sine and Cosine

Examples

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