Question

If `f(x)= 9sin(sinx^2)`, then what is `f^'(x)`?

Answer 1

`f^'(x) = 18x * cosx^2 *cos(sinx^2)`

Explanation:

Start by breaking down your function - this will give you an idea about what you need to use to differentiate it

`f(x) = y = 9 * sin(sin(x^2))`

From the looks of it, you're going to need to use the chain rule twice to differentiate `sin(sin(x^2))`.

The chain rule tells you that you can differentiate a function that is actually the composition of two other functions like this

`color(blue)(d/dx(y) = d/(du)(y) * d/dx(u))`, where

`y` is a function that depends on variable `u`, which in turn depends on variable `x`.

Another important thing to remember is that

`d/dx(sinx) = cosx`

So, you have `sin (sinx^2)` as part of your original function. First, take `sin(x^2)` to be `u` so that you have

`y = 9 * sinu`

This can be differentiated like this

`d/dx(y) = d/(du)(9 * sinu) * d/dx(u)`

`d/dx(y) = 9 * d/(du)(sinu) * d/dx(u)`

Now you need to use the chain rule for a second time, since you have

`u = sinx^2`

This time, you can write `u = sinu_1`, where `u_1 = x^2`. This will get you

`d/dxu = d/(du_1) * sinu_1 * d/(dx)u_1`

`d/dx(u) = cosu_1 * 2x`, which is equivalent to

`d/dx(u) = cosx^2 * 2x`

Your original derivative now becomes

`d/dx(y) = 9 * cos(sinx^2) * 2x * cosx^2`

which is equal to

`d/dx(y) = color(green)(18x * cosx^2 * cos(sinx^2))`