Does anyone want to help me with calculus? If I don't figure this out tonight I shall fail, and my usual math help has temporarily died.
f(x) = cos³(sin2x)
3cos²u · u'
3cos²(sin2x) · ????
*** I know you use the chain rule...
So you would find the derivative of sin(2x), which means you have to do the chain rule again?
So that's 2cos2x?
I have the answer to the problem and there's a whole other
-sin(sin2x) term that somehow worked its way in there.
How do you even begin f(x) = [x sin2x + tan^4(x^7)]^5?
Updated On: 5/24/06 at 08:36 PM