- #1

- 23

- 0

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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- Thread starter staka
- Start date

- #1

- 23

- 0

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.

- #2

- 102

- 0

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

- #3

- 23

- 0

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

- #4

Mark44

Mentor

- 36,042

- 7,977

You need the derivative of cos

- #5

cristo

Staff Emeritus

Science Advisor

- 8,122

- 74

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.

- #6

- 23

- 0

Alright. I'll retry.. solving.

- #7

- 102

- 0

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

- #8

- 23

- 0

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

- #9

- 23

- 0

Alright thanks. I've solved it correctly.

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