## anonymous 5 years ago Find dy/dx for: y = cos^2(sinx)

1. myininaya

y'=(sinx)'2*cos(sinx)(cos)'(sinx)

2. anonymous

hey is it y= cos$(sinx)^{2}$

3. myininaya

y'=(cosx)2[cos(sinx)](-sin(sinx))

4. anonymous

Huh? $\frac{dy}{dx} = 2cos(sin\ x)(-sin(sin\ x))(cos\ x)$

5. anonymous

Whole lot of chain rule goin on.

6. myininaya

it might be easier if you rewrite y=[cos(sinx)]^2

7. anonymous

Can simplify to $-2cos^2(sin\ x)*sin(sin\ x)$ or $2(sin^3(sin\ x)) - sin(sin\ x))$ if you prefer only dealing with signs.

8. anonymous

Err sines.

9. anonymous

Sine sine everywhere a sine..

10. anonymous

ok so those are simplified versions. So to find the derivative of it, you product rule the first one (somehow). And the second one, you do product rule for the two terms?

11. anonymous

No, it's just chain rule. First we take the derivative of $$cos^2(sin\ x)$$ just as if it were $$u^2$$

12. anonymous

Those equations look crazy confusing. I have this question on my test tomorrow, and theres about a 2 inches of blank space to find it. Is there an easier way?

13. anonymous

We get 2u*(the derivative of u)

14. anonymous

Where u is the cos(sin x)

15. anonymous

So then to find the derivative of u we take the derivative of the cos(sin x) just as if it were cos p. And we get -sin(p)*(derivative of p)

16. anonymous

Where p is sin x. So the derivative of p is the derivative of sin x. Which is cos x.

17. anonymous

Does that make sense?

18. anonymous

I'm trying it right now. YOU ARE GETTING A MEDAL NO MATTER WHAT :D

19. anonymous

So dy/dx = 2(-sin(sinx))(cosx)?

20. anonymous

Close, you're missing a cos(sin x)

21. anonymous

Actually my second post with the 'simplifications' are totally wrong for that reason ;p But the first version I gave was correct.

22. anonymous

Wait, isn't it: 2(cos(sinx))(-sinx(sinx))(cosx) (I forgot the 2nd part of the chain rule >_>)

23. anonymous

and yes it is :D