# 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
Correct, instead of:

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

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

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