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

2 Answers
Mar 27, 2015

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

Mar 27, 2015

"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)
}