Question

Find dy/dx for: y = cos^2(sinx)

Answer 1

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

Answer 2

hey is it y= cos\[(sinx)^{2}\]

Answer 3

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

Answer 4

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

Answer 5

Whole lot of chain rule goin on.

Answer 6

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

Answer 7

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.

Answer 8

Err sines.

Answer 9

Sine sine everywhere a sine..

Answer 10

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?

Answer 11

No, it's just chain rule.

First we take the derivative of \(cos^2(sin\ x)\) just as if it were \(u^2\)

Answer 12

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?

Answer 13

We get 2u*(the derivative of u)

Answer 14

Where u is the cos(sin x)

Answer 15

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)

Answer 16

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

Answer 17

Does that make sense?

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

Actually my second post with the 'simplifications' are totally wrong for that reason ;p

But the first version I gave was correct.

Answer 22

Wait, isn't it:

2(cos(sinx))(-sinx(sinx))(cosx)

(I forgot the 2nd part of the chain rule >_>)

Answer 23

and yes it is :D