1. ## Derivative Help

find the derivative:
s(x)=sin^2⁡((3x-pi))

could someone show me how to find the derivative of this using both methods of the chain rule: F'(x) = f '(g(x)) g '(x) and (dy/du)*(du/dx).
I keep getting 2(sin(3x-pi)(-3cos(3x)) which i know to be incorrect

2. Originally Posted by ~berserk
find the derivative:
s(x)=sin^2⁡((3x-pi))

could someone show me how to find the derivative of this using both methods of the chain rule: F'(x) = f '(g(x)) g '(x) and (dy/du)*(du/dx).
I keep getting 2(sin(3x-pi)(-3cos(3x)) which i know to be incorrect
$s=u^2 \quad u=\sin(y) \quad y=3x-\pi$

Now by the chain rule we get

$\frac{ds}{dx}=\frac{ds}{du}\frac{du}{dy}\frac{dy}{ dx}=2u\cos(y)(3)=6\sin(y)\cos(y)=6\sin(3x-\pi)\cos(3x-\pi)$

3. Is there any way you could show your process in a little more detail?

4. Originally Posted by ~berserk
Is there any way you could show your process in a little more detail?
$\frac{ds}{dx}=\underbrace{\frac{ds}{du}}_{2u} \underbrace{\frac{du}{dy}}_{\cos(y)} \underbrace{ \frac{dy}{dx} }_{3}=2u\cos(y)(3)$

Now we need to back substitute the variable in the first post

$6u\cos(y)=6\sin(y)\cos(y)=6\sin(3x-\pi)\cos(3x-\pi)$