Question

What is `f'(-pi/3)` when you are given `f(x)=sin^7(x)`?

Answer 1

It is `(7sqrt3)/2^7=(7sqrt3)/128`

Method

`f(x)=sin^7(x)`

It is very useful to re-write this as `f(x)=(sin(x))^7` because this makes it clear that what we have is a `7^(th)` power function.

Use the power rule and the chain rule (This combination is often called the generalized power rule.)

For `f(x)=(g(x))^n`, the derivative is `f'(x)=n(g(x))^(n-1)*g'(x)`,

In other notation `d/(dx)(u^n)=n u^(n-1) (du)/(dx)`

In either case, for your question `f'(x)=7(sin(x))^6*cos(x)`

You could write `f'(x)=7sin^6(x) *cos (x)`

At `x=- pi/3`, we have
`f'(- pi/3)=7sin^6(- pi/3) *cos (- pi/3)=7(1/2)^6(sqrt3/2)=(7sqrt3)/2^7`

Answer 2

`"let " y= f(x)` `=> dy/dx = f'(x)`

`=>y = sin^7(x)`

`"let " u = sin(x) => y = u^7`

`du/dx = cos(x)`

`dy/du = 7*u^6`

Now, `f'(x) = (dy)/(dx)`
`= (dy)/(du)*(du)/(dx)` {Do you agree?}
`= 7u^6*cosx`
but remember `u = sin(x)`
`=> f'(x) = 7sin^6(x)cos(x)`

`=> f'(-pi/3) = 7*(sin(-pi/3))^6**cos(-pi/3)`

`= 7(-sqrt(3)/2)^6**(1/2)`

You Have The Honor To Simplify

NOTE:
{
wondering why im doing all this "let stuff" ?

the reason is there are more than one function in `f(x)`

** there's : `sin^7(x)` and there's `sin(x)`!!

so to find the `f'(x)` i need to find the `f'` of `sin^7(x)`
AND the `f'` of `sin(x)`

that's why i need to let `y= f(x)`
then let `u = sin(x)`
}