# D/dx of Inverse trig

1. Jan 29, 2006

### Chocolaty

How do you derive a function where the trig is under a square root
like this:
y=sqr(tan^-1(x))
y=(tan^-1(x))^(1/2)

i know that y=tan^-1u(x) => dy/dx=1/(1+u^2)*du/dx

But how do I work it out? The book doesn't give the answers to pair numbers :(

2. Jan 29, 2006

### StatusX

chain rule?

3. Jan 29, 2006

### d_leet

Yea that's what I was going to say.

4. Jan 29, 2006

### Jameson

Yeah, chain rule.

$$u=\tan^{-1}(x)$$
$$du=\frac{dx}{x^2+1}$$

So using u-substitution you can write you function as this.

$$y=\sqrt{u}=u^{\frac{1}{2}}$$
$$y'=\frac{1}{2\sqrt{u}}*du$$

I think you can make the neccessary substitution now.

Last edited: Jan 29, 2006
5. Jan 29, 2006

### VietDao29

The second line is wrong, Jameson.
It should read:
$$du=\frac{dx}{x^2+1}$$.
dx, not 1 .
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Have you worked out the problem, Chocolaty?

Last edited: Jan 29, 2006
6. Jan 29, 2006

### d_leet

Yes I have.

7. Jan 29, 2006

### VietDao29

Ooops, sorry, I mean Chocolaty ...
Hope you don't mind.

8. Jan 29, 2006

### d_leet

lol That's alright.

9. Jan 29, 2006

### Jameson

Tsk tsk. Sorry bout that. Fixed.

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