Derivative involving inverse trigonometric functions

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Homework Help Overview

The discussion revolves around finding the derivative of a function involving a square root and an inverse trigonometric function, specifically the derivative of sqrt(x^2-4) - 2tan^-1{0.5*sqrt(x^2-4)}.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the components of the derivative, questioning the correctness of the expressions for U' and 1+U^2. There is an exploration of potential errors in calculations and the interpretation of the function.

Discussion Status

Some participants have offered clarifications regarding the notation and components involved in the derivative. There is an acknowledgment of mistakes in calculations, and some participants are working through their reasoning to identify errors.

Contextual Notes

There are indications of confusion regarding the proper formulation of the derivative and the components involved, particularly concerning the expressions for U and its derivative. Participants are also reflecting on their understanding of the function's structure.

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
 
Last edited:
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!
 

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