Deriving D/dx of Inverse Trig Functions

[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 :(

 

[StatusX]

chain rule?

 

[d_leet]

chain rule?

Yea that's what I was going to say.

 

[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 necessary substitution now.

 

[VietDao29]

Yeah, chain rule.

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

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?

 

[d_leet]

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Have you worked out the problem, d_leet?

Yes I have.

 

[VietDao29]

Yes I have.

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

 

[d_leet]

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

lol That's alright.

 

[Jameson]

The second line is wrong, Jameson.
It should read:
du=\frac{dx}{x^2+1}.
dx, not 1 .

Tsk tsk. Sorry bout that. Fixed.