## anonymous 5 years ago find the derivative of y=sin^3 x

1. anonymous

What you want to remember here is the chain rule. Take each part by itself. =sin^3(x) = 3sin^2(x) = 3cos^2(x)*1 = 3cos^2(x) Remember the 1 from the inner x, if it's something else then x it could change the result. Unless I'm forgetting the chain rule myself, but it should be correct. :)

2. anonymous

THANK YOU :)

3. anonymous

You're welcome :)

4. anonymous

This is incorrect. The inner function is $f(x) = \sin{x}$ and the outer function is $g(x) = x^{3}$ Thus $\frac{d}{dx} \sin^3{x} = \frac{d}{dx} g \circ f = \frac{dg}{dx}(f(x)) \cdot \frac{df}{dx} = 3\sin^2{x} \cdot \cos{x}$

5. anonymous

Yeah, thought I did something wrong. Good catch!