Question

What are the critical points of `f(x) = (x - 3sinx)^4 -xcosx`?

Answer 1

Critical Points are where the derivative is equal to 0.

But we need to find the derivative. In order to do so, I'll split up `f(x)` into two parts -PART 1 = `(x-3sin(x))^4` and PART 2 = `xcos(x)` and I will combine them using the Sum Rule later.

PART 1:

Let's apply the Chain Rule.

`d/dx(x-3sin(x))^4=4*(x-3sin(x))^3*(1-3cos(x))`
`=4(x-3sin(x))^3(1-3cos(x))`

PART 2:

Let's apply the Product Rule.

`d/dx xcos(x)=1*cos(x)+x*-sin(x)`
`=cos(x)-xsin(x)`

`therefore f'(x)` is:

`=4(x-3sin(x))^3(1-3cos(x))-(cos(x)-xsin(x))`
`=xsin(x)-cos(x)+4(x-3sin(x))^3(1-3cos(x))` (Rearranging and expanding `-(cos(x)-xsin(x))`)

Now all you have to do is set that equal to 0:

`xsin(x)-cos(x)+4(x-3sin(x))^3(1-3cos(x))=0`

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