The Added Constants and Multiplied Constants: Their Effects on Derivatives

Example 5.1. Let $$y=x^3+5.$$ Just as before, let us suppose $x$ to grow to $x+dx$ and $y$ to grow to $y+dy$.

Then: \begin{align} y + dy &= (x + dx)^3 + 5 \\ &= x^3 + 3x^2\, dx + 3x(dx)^2 + (dx)^3 + 5. \end{align} Neglecting the small quantities of higher orders, this becomes $$y+dy=x^3+3x^2\cdot dx+5.$$ Subtract the original $y=x^3+5$, and we have left: \begin{align} dy &= 3x^2\, dx. \\ \frac{dy}{dx} &= 3x^2. \end{align}

So the $5$ has quite disappeared. It added nothing to the growth of $x$, and does not enter into the derivative. If we had put $7$, or $700$, or any other number, instead of $5$, it would have disappeared. So if we take the letter $a$, or $b$, or $c$ to represent any constant, it will simply disappear when we differentiate.

Example 5.2. Let $y=7x^2$. 
Then on proceeding as before we get: \begin{align} y + dy &= 7(x+dx)^2 \\ &= 7\left\{x^2 + 2x\cdot dx + (dx)^2\right\} \\ &= 7x^2 + 14x\cdot dx + 7(dx)^2. \end{align} Then, subtracting the original $y=7x^2$, and neglecting the last term, we have \begin{align} dy &= 14x\cdot dx.\\ \frac{dy}{dx} &= 14x. \end{align}

Example 5.3. Differentiate $y={\displaystyle \frac{x^5}{7}}-{\displaystyle \frac{3}{5}}$.

Solution. ${\displaystyle \frac{3}{5}}$ is an added constant and vanishes (see here).

We may then write at once $$\frac{dy}{dx}=\frac{1}{7}\times 5\times x^{5-1},$$ or $$\frac{dy}{dx}=\frac{5}{7}{x}^4.$$

Example 5.4. Differentiate $y=a\sqrt{x}-{\displaystyle \frac{1}{2}}\sqrt{a}$.

Solution. The term ${\displaystyle \frac{1}{2}}\sqrt{a}$ vanishes, being an added constant; and as $a\sqrt{x}$, in the index form, is written $a{x}^{\frac{1}{2}}$, we have $$\frac{dy}{dx}=a\times \frac{1}{2}\times x^{\frac{1}{2}-1}=\frac{a}{2}\times x^{-\frac{1}{2}},$$ or $$\frac{dy}{dx}=\frac{a}{2\sqrt{x}}.$$

Example 5.5. If $$ay+bx=by-ax+(x+y)\sqrt{a^2-b^2},$$ find the derivative of $y$ with respect to $x$.

Solution. As a rule an expression of this kind will need a little more knowledge than we have acquired so far; it is, however, always worth while to try whether the expression can be put in a simpler form.

First, we must try to bring it into the form $y=$ some expression involving $x$ only.

The expression may be written $$(a-b)y+(a+b)x=(x+y)\sqrt{a^2-b^2}.$$

Squaring, we get $$(a-b{)}^2{y}^2+(a+b{)}^2{x}^2+2(a+b)(a-b)xy=(x^2+y^2+2xy)(a^2-b^2),$$ which simplifies to $$(a-b{)}^2{y}^2+(a+b{)}^2{x}^2=x^2(a^2-b^2)+y^2(a^2-b^2);$$ or $$[(a-b{)}^2-(a^2-b^2)]y^2=[(a^2-b^2)-(a+b{)}^2]x^2,$$ that is $$2b(b-a)y^2=-2b(b+a)x^2;$$

hence² $$y=\sqrt{\frac{a+b}{a-b}}x\text{and}\frac{dy}{dx}=\sqrt{\frac{a+b}{a-b}}.$$

Example 5.6. (a) The volume of a cylinder of radius $r$ and height $h$ is given by the formula $V=\pi r^{2}h$. Find the rate of variation of volume with the radius when $r=5.5$ in and $h=20$ in (b) If $r=h$, find the dimensions of the cylinder so that a change of $1$ in in radius causes a change of $400$ in${}^3$ in the volume.

Solution. (a) The rate of variation of $V$ with regard to $r$ is $$\frac{dV}{dr}=2\pi rh.$$

If $r=5.5$ in and $h=20$ in, this becomes $690.8$. It means that a change of radius of $1$ inch will cause a change of volume of $690.8$ in${}^3$. This can be easily verified, for the volumes with $r=5$ and $r=6$ are $1570$ in${}^3$ and $2260.8$ in${}^3$ respectively, and $2260.8-1570=690.8$.

(b) If $$r=h,\text{and}\frac{dV}{dr}=2\pi r^2=400\frac{{\text{in}}^3}{\text{in}}$$ we must have $$r=h=\sqrt{\frac{400}{2\pi }}\approx 7.98\text{in}.$$

Example 5.7. The reading $\theta$ of a Féry’s Radiation pyrometer is related to the Centigrade temperature $T$ of the observed body by the relation $$\frac{\theta }{{\theta }_1}={\left(\frac{T}{T_1}\right)}^4,$$ where ${\theta }_1$ is the reading corresponding to a known temperature $T_1$ of the observed body.

Compare the sensitiveness of the pyrometer at temperatures $800{}^{\circ }$C, $1000{}^{\circ }$C, $1200{}^{\circ }$C, given that it read $25$ when the temperature was $1000{}^{\circ }$C.

Solution. The sensitiveness is the rate of variation of the reading with the temperature, that is ${\displaystyle \frac{d\theta }{dT}}$. The formula may be written $$\theta =\frac{{\theta }_1}{T_1^4}{T}^4=\frac{25T^4}{{1000}^4},$$ and we have $$\frac{d\theta }{dT}=\frac{100T^3}{{1000}^4}=\frac{T^3}{10,000,000,000}.$$

When $T=800{}^{\circ }$C, $1000{}^{\circ }$C and $1200{}^{\circ }$C, we get ${\displaystyle \frac{d\theta }{dT}}=0.0512$, $0.1$ and $0.1728$, respectively. The sensitiveness is approximately doubled from $800{}^{\circ }$C to $1000{}^{\circ }$C, and becomes three-quarters as great again up to $1200{}^{\circ }$C.