Calculus Volume 1

296

Chapter 3 | Derivatives

du dx = 2

(3 x +2) 2

and

dy du =5

u 4 .

Finally, we put it all together. dy dx = dy du ·

du dx

Apply the chain rule.

dy du =5

u 4 and du

2 (3 x +2) 2

=5 u 4 ·

Substitute

dx = 2

(3 x +2) 2 .

⎛ ⎝ x

⎞ ⎠

4

x 3 x +2 .

2 (3 x +2) 2

=5

·

Substitute u =

3 x +2

4 (3 x +2) 6

= 10 x

Simplify.

It is important to remember that, when using the Leibniz form of the chain rule, the final answer must be expressed entirely in terms of the original variable given in the problem.

Example 3.59 Taking a Derivative Using Leibniz’s Notation, Example 2

Find the derivative of y = tan ⎛

⎝ 4 x 2 −3 x +1 ⎞ ⎠ .

Solution First, let u =4 x 2 −3 x +1. Then y = tan u . Next, find du dx and dy du : du dx =8 x −3and dy du = sec 2 u . Finally, we put it all together. dy dx = dy du · du dx

Apply the chain rule.

dy du = sec

= sec 2 u · (8 x −3) 2 u . = sec 2 (4 x 2 −3 x +1) · (8 x − 3) Substitute u =4 x 2 −3 x +1. Use du dx =8 x −3and

Use Leibniz’s notation to find the derivative of y =cos ⎛ ⎝ x 3 ⎞

3.41

⎠ . Make sure that the final answer is

expressed entirely in terms of the variable x .

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