# How may I derive this equation?

How may I derive y=cos^3(sinx)?
I thought simply just y=3cos(sinx)^2(cosx)...

Answer given is -3cosxsin(sinx)cos^2(sinx) but no idea how to get to that.

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C.E
I am a bit confused about what the exact question is. What is it?

How to derive the function: y=cos^3(sinx) is my question... solving for y'=.

Mark44
Mentor
I think you didn't use the chain rule enough times.
You need the derivative of cos3(sin(x)) with respect to cos(sin(x)) times the derivative of cos(sin(x)) wrt sin(x) times the derivative of sin(x) wrt x.

cristo
Staff Emeritus
By derive you mean differentiate, right?

Try writing t=sinx, then y=[cos(t)]^3. Then, dy/dx=(dy/dt)(dt/dx). Course, you need to use the chain rule again when differentiating y wrt t.

Alright. I'll retry.. solving.

C.E
Oh, differentiate now I know what you are trying to do. Are you familiar with the chain rule?

I am familar with the chain rule.. I'll retry.

Alright thanks. I've solved it correctly.