Derivative involving inverse trigonometric functions

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biochem850
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Homework Statement


Find the derivative of:
sqrt(x^2-4)-2tan^-1{.5*sqrt(x^2-4)}




Homework Equations


U'/1+U^2
U'=x/2sqrt(x^2-4)
1+U^2=x^2


The Attempt at a Solution



I combined the above components but my answer is incorrect. I feel that I might have the wrong answer for "1+U^2". I just cannot seem to catch my error at the moment.
 
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Did you mean the following function? [tex]\sqrt{(x^2-4)}-2\arctan\left(0.5 \sqrt{(x^2-4)}\right)[/tex]

What do you mean on U'/1+U^2? Are no parentheses missing?

Did you mean the derivative of arctan(U)?

[tex]\frac{U'}{1+U^2}[/tex]

with [tex]U=0.5\sqrt{(x^2-4)}[/tex]
and
[tex]U'=0.5 \frac{x}{\sqrt{(x^2-4)}}[/tex]?Recalculate U^2+1. It is not x^2.

ehild
 
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My U'=[itex]\frac{x}{2\sqrt{x^2-4}}[/itex]

When I square [itex]\frac{\sqrt{x^2-4}}{2}[/itex] and add one the only other answer I get is [itex]\frac{x^2}{4}[/itex]
I've been working for a while and perhaps I'm missing something very simple.
 
Last edited:
I've gotten the answer:

[itex]\frac{\sqrt{x^2-4}}{x}[/itex]

I made a simple mistake in calculating 1+U[itex]^{2}[/itex]

Thanks for the help!