Question

How do you use the chain rule to differentiate `y=cos((4x)^3)`?

Answer 1

`y=cos((4x)^3) => y'=-192x^2sin((4x)^3)`

Explanation:

The Chain Rule States that

`(f(g(x))'=f'(g(x))(g'(x))`

in the case of `y=cos((4x)^3)`

`f(x)=cos(x)`

and

`g(x)=(4x)^3`

`f(x)=cos(x) => f'(x)=-sin(x)`

and `f'(g(x))=-sin((4x)^3)`

Also since `(ab)^x=a^x(b^x)`

then `(4x)^3=4^3(x^3)=64x^3`

And since `f'(cx)=c(f'(x))`

`g'(x)=(64x^3)=64(x^3)'`

and since the power rule states that `(x^n)'=nx^(n-1)`

`g'(x)=192x^2`

Then we plug in and get

`y'=-192x^2sin((4x)^3)`

Answer 2

`d/(dx) [cos((4x)^3)] = color(blue)(-192x^2·sin(64x^3)`

Explanation:

We're asked to find the derivative

`d/(dx) [cos((4x)^3)]`

which is also

`d/(dx) [cos(64x^3)]`

Using the chain rule, which for this situation is

`d/(dx)[cos(64x^3)] = d/(du) [cosu] (du)/(dx)`

where

• `u = 64x^3`

• `d/(du) [cosu] = -sinu`:

`= -sin(64x^3)·d/(dx)[64x^3]`

Using the power rule on the `64x^3` term:

`d/(dx) [64x^3] = 3·64x^(3-1)`

`= color(blue)(ul(-192x^2·sin(64x^3)`