Chain Rule

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GRE Quantitative Reasoning › Chain Rule

Questions 1 - 3

1

Compute the derivative:

Explanation

This question requires application of multiple chain rules. There are 2 inner functions in , which are and .

The brackets are to identify the functions within the function where the chain rule must be applied.

Solve the derivative.

$\frac{dy}{dx}= -sin([sin(2x)]) \cdot (cos[2x]))\cdot 2=-2cos(2x)sin(sin(2x))$

The sine of sine of an angle cannot be combined to be sine squared.

Therefore, the answer is:

2

What is the derivative of

Explanation

Chain Rule:

$\frac{\mathrm{dy} }{\mathrm{d} x}=f(x)g'(y)+g(y)f'(x)$

For this problem

$f'(x)=\frac{\mathrm{d} }{\mathrm{d} x}3x^2=3(2)x^{2-1}=6x$

$g'(y)=\frac{\mathrm{d} }{\mathrm{d} x}8y^3=8(3)y^{3-1}=24y^2$

Plug the values into the Chain Rule formula and simplify:

$\frac{\mathrm{dy} }{\mathrm{d} x}(3x^2*8y^3)=3x^2*24y^2+8y^3*6x$

3

Find the derivative of the following function:

$h(x)=(3x^3-4x)^{33}$

$h'(x)=(297x^2-132)(3x^2-4x)^{32}$

$h'(x)=(297x^2-132)(3x^2-4x)^{33}$

$h'(x)=(297x^2+132)(3x^2-4x)^{33}$

$h'(x)=(297x^2)(3x^2-4x)^{32}$

$h'(x)=(9x^2-4)(3x^2-4x)^{32}$

Explanation

Recall chain rule for this problem

So if we are given the following,

$h(x)=(3x^3-4x)^{33}$

We can think of it like this

$g(x)=x^{33}$

$g'(x)=33x^{32}$

$(f(g(x)))'=(9x^2-4)(33)(3x^2-4x)^{32}$

Clean it up a bit to get:

$(f(g(x)))'=(297x^2-132)(3x^2-4x)^{32}$

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