Chain Rule - GRE Subject Test: Math

Card 0 of 3

Question

Compute the derivative:

Answer

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:

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Question

Find the derivative of the following function:

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

Answer

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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Question

What is the derivative of

Answer

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:

$\frac{\mathrm{dy} }{\mathrm{d} x}(3x^28y^3)=3x^224y^2+8y^3*6x$

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