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 StevenMx February 2nd, 2009 04:09 PM

Problem with derivative

I would be really grateful if anyone could help me with this derivative:

$y=\frac{arctan2x}{arccot2x}$

OK, I believe I need to use the quotient rule, and I'll get something like that:

$\frac{(arctan2x)'arccot2x \cdot arctan2x(arccot2x)'}{(arccot2x) ^{2} }$

but I don't know what to do next...

 StevenMx February 3rd, 2009 03:20 AM

Re: Problem with derivative

Updated my previous post :)

 CogitoErgoCogitoSum February 3rd, 2009 09:46 AM

Re: Problem with derivative

Quote:
 Originally Posted by StevenMx $\frac{(arctan2x)'arccot2x \cdot arctan2x(arccot2x)'}{(arccot2x) ^{2} }$ but I don't know what to do next...
First of all, your quotient rule is wrong. You add, not multiply, the two terms
$\frac{\arctan(2x)'\cdot arccot(2x) + \arctan(2x)\cdot arccot(2x)'}{arccot^2(2x)}$

Now, from trigonometry, we ought to know that:
$arccot\; r= \arctan\frac{1}{r}$

$\frac{\arctan(2x)'\cdot \arctan(\frac{1}{2x}) + \arctan(2x)\cdot \arctan(\frac{1}{2x})'}{\arctan^2(\frac{1}{2x} )}$

The real question is this... do you know how to find the derivative of an inverse function?

Follow this procedure for any arbitrary inverse function...

Suppose $y= f^{-1}(x)$

Then $f(y)= x$ by definition of an inverse function

Find the derivative: $f'(y)\cdot y' = 1$ utilizing both chain rule and implicit differentiation.

Solve for y': $y'= \frac{1}{f#39;(y)}$

Plug in your function y in terms of x:
$y'= \frac{1}{f#39;(f^{-1}(x))}$

There... thats all you need to know for an arbitrary f(x)... you can find the derivative of any f^-1.

$y= f^{-1}(x) = \arctan x$,

we know that
$\tan y= x$ and therefore

$\sec^2 y \cdot y' = 1$

$y' = \cos^2 y$

$y' = \cos^2 \arctan(x)$

$\frac{d}{dx} \arctan(x)= \cos^2(\arctan(x))$

Refer back to your original problem:
$\frac{\arctan(2x)'\cdot \arctan(\frac{1}{2x}) + \arctan(2x)\cdot \arctan(\frac{1}{2x})'}{\arctan^2(\frac{1}{2x} )}$

$\frac{2\cos^2(\arctan(x))\cdot \arctan(\frac{1}{2x}) - \frac{1}{2x^2}\arctan(2x)\cdot \cos^2(\arctan(\frac{1}{2x}))}{\arctan^2(\frac{1}{ 2x})}$

You have to look back at good old trigonometry yet again...
$\cos \; \arctan \;r= \frac{1}{\sqrt{1 + r^2}}$
and so
$\cos^2 \; \arctan \;r= \frac{1}{1 + r^2}$

Plugging in:
$\frac{2\frac{1}{1 + x^2}\cdot \arctan(\frac{1}{2x}) - \frac{1}{2x^2}\arctan(2x)\cdot \frac{1}{1 + (\frac{1}{2x})^2} }{\arctan^2(\frac{1}{2x})}$

$\frac{\frac{2}{1 + x^2}\cdot \arctan(\frac{1}{2x}) - \arctan(2x)\cdot \frac{2}{4x^2 + 1} }{\arctan^2(\frac{1}{2x})}$

This problem is boring me. I will quit now.

 StevenMx February 4th, 2009 03:18 AM

Re: Problem with derivative

Thanks man :) I really appreciate that.

 shynthriir February 4th, 2009 04:58 AM

Re: Problem with derivative

Quote:
 First of all, your quotient rule is wrong. You add, not multiply, the two terms

d/dx ( u / v) = (u' * v - u * v') / v^2

 CogitoErgoCogitoSum February 4th, 2009 06:56 AM

Re: Problem with derivative

Quote:

Originally Posted by shynthriir
Quote:
 First of all, your quotient rule is wrong. You add, not multiply, the two terms

d/dx ( u / v) = (u' * v - u * v') / v^2

LOL. you certainly do. Im sorry man, I totally screwed that up for you. Just change the signs.

 StevenMx February 4th, 2009 07:14 AM

Re: Problem with derivative

Don't worry man, nothing happened :) I changed the signs.

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