Homework Help: Derivative using chain rule

1. Jul 5, 2008

jkeatin

1. The problem statement, all variables and given/known data
y= squareroot tan(sin^2 x)

2. Relevant equations
chain rule

3. The attempt at a solution
f(x)= sqaureroot tan x
g(x)= (sinx)^2
f'(x)=1/2 sec^2x ^1/2
g'(x)= 2 * sinx * cosx

I dont know if my f'(x) is right if it is then do i just do the chain rule?

2. Jul 5, 2008

rocomath

$$F(x)=\sqrt{\tan{\sin^2 x}}$$

$$f(x)=\sqrt x$$
$$g(x)=\tan x$$
$$h(x)=\sin^2 x$$

$$F(x)=f\{g[h(x)]\}$$

3. Jul 5, 2008

Hootenanny

Staff Emeritus
I'm afraid your f'(x) isn't correct, your g'(x) however is. There is no harm in using the chain rule one more time to make life a little easier. For example, let:

$$g(x) = \sin^2x$$

$$f(g) = \tan\left(g\right)$$

$$h(f) = \sqrt{f}$$

Then,

$$\frac{d}{dx}\sqrt{\tan\left(\sin^2x\right)} = \frac{dh}{df}\frac{df}{dg}\frac{dg}{dx}$$

Do you follow?

Edit: It seems I was beaten to it.

Last edited: Jul 5, 2008
4. Jul 5, 2008

jkeatin

yeah kinda

1/2x(-1/2)tan(sin^2x)[sec^2x(sin^2x)](2sinxcosx)

is that makin any progress?

5. Jul 5, 2008

Hootenanny

Staff Emeritus
You're not far off. Are you sure that x should be an x?

6. Jul 5, 2008

jkeatin

is it just 1/2tan(sin^2x)

i just thought because f'(x)= 1/2x-1/2

7. Jul 5, 2008

Hootenanny

Staff Emeritus

$$\frac{1}{2}x^{-1/2}\ldots$$

it should be

$$\frac{1}{2}\tan^{-1/2}\left(\sin^2x\right)\ldots$$
No it isn't.

Last edited: Jul 5, 2008
8. Jul 5, 2008

jkeatin

1/2tan^1/2(sin^2x)[sec^2x(sin^2x)](2sinxcosx)
thats it?

9. Jul 6, 2008

Hootenanny

Staff Emeritus
Careful with your exponent of the tangent, and don't forget that you can cancel the 1/2 with the 2.

10. Jul 6, 2008

jkeatin

tan^-1/2(sinx)...

11. Jul 6, 2008

jkeatin

or you mean the 2 in 2sinxcosx

12. Jul 6, 2008

jkeatin

tan^-1/2(sin^2x)[sec^2x(sin^2x)](sinxcosx)

13. Jul 6, 2008

Hootenanny

Staff Emeritus
Spot on

14. Jul 6, 2008

jkeatin

thanks for the help dude

15. Jul 6, 2008

Hootenanny

Staff Emeritus
No problem