Sometimes one is stumped by finding that the expression to be differentiated is too complicated to tackle directly.
Thus, the equation
Now the dodge to turn the difficulty is this: Write some symbol, such as
for
By and bye, when you have learned how to deal with sines, and cosines, and exponentials, you will find the chain rule of increasing usefulness.
Examples
Let us practice using the Chain Rule on a few examples.
Example 9.1. Differentiate
Solution. Let
Example 9.2. Differentiate
Solution. Let
Example 9.3. Differentiate
Solution. Let
Example 9.4. Differentiate
Solution. Let
Example 9.5. Differentiate
Solution. Write this as
(We may also write
Proceeding as in example 9.1 above, we get
Hence
Example 9.6. Differentiate
Solution. We may write this
Differentiating
Example 9.7. Differentiate
Solution. Let
Now let
Example 9.8. Differentiate
Solution. We get
Let
Let
Hence
Example 9.9. Differentiate
Solution.
Example 9.10. Find the first and second derivatives of
Solution.
Let
Hence
Now
(We shall need these two last derivatives later on. See exercise 11 from Chapter 12.)
Example 9.11. A cylinder whose height is twice the radius of the base is increasing in volume, so that all its parts keep always in the same proportion to each other; that is, at any instant, the cylinder is similar to the original cylinder. When the radius of the base is
Solution.
The volume changes at the rate of
Exercises I
Differentiate the following:
Exercise 9.1.
Answer
Solution
Let
Exercise 9.2.
Answer
Solution
Let
Exercise 9.3.
Answer
Solution
Exercise 9.4.
Answer
Solution
Let
and
Exercise 9.5.
Answer
Solution
To find
Exercise 9.6.
Answer
Solution
To find
Now using the Quotient Rule:
Exercise 9.7.
Answer
Solution
Using the Quotient Rule:
Note that to find
Exercise 9.8. Differentiate
Answer
Solution
Exercise 9.9. Differentiate
Answer
Solution
To find
Exercise 9.10. A spherical balloon is increasing in volume. If, when its radius is
Answer
At the rate of
Solution
The volume of the balloon is
and the surface of the balloon is
We know
We want to find
Differentiate both sides of the following equation with respect to time
Since
Now differentiate both sides of
Substitute
The process can be extended to three or more derivatives, so that
Examples
Example 9.12. If
Solution. We have
Example 9.13. If
Solution. Since
Differentiating
So
Example 9.14. If
Solution. We get
So that
Replace now first
Exercises II
Exercise 9.11. If
Answer
Solution
Exercise 9.12. If
Answer
Solution
Exercise 9.13. If
Answer
Solution