Question

How do you differentiate `f(x)=(sin2x^2)/4` using the chain rule?

Answer 1

`f'(x)=xcos(2x^2)`

Explanation:

According to the chain rule, since the derivative of `sin(x)` is `cos(x)`, we will differentiate the outside `sin` function to give us `cos`, and then multiply that by the derivative of the inside function, which is `2x^2`.

Formally, this can be written as

`d/dx[sin(g(x))]=cos(g(x))*g'(x)`

Here, notice first that `sin(2x^2)` is being multiplied by `1"/"4`, which we can bring out of the expression:

`f'(x)=1/4cos(2x^2)*d/dx[2x^2]`

Since the derivative of `2x^2` is `4x`, this gives us

`f'(x)=1/4cos(2x^2)*4x`

The `4`s will cancel, leaving us with

`f'(x)=xcos(2x^2)`