# Derivative problem

• Sep 17th 2008, 05:34 PM
NotEinstein
Derivative problem
$sqrt((cot^2(x))+x^2)$

My attempt-

$sqrt(2cot(x) * (-csc^2x) +2x)
$

Here I believe I have two options.. not sure which one to go with if either..
$sqrt(2cot(x) * (-cot^2x - 1) +2x)
$

or

$sqrt(2cot(x) * (1/sin^2x) +2x) = sqrt(2cosx/sinx * (1/sin^2x)+2x)$

just trying to get it to simplest terms.
• Sep 17th 2008, 05:44 PM
skeeter
do you know how the chain rule works?

$\frac{d}{dx} \left[(\cot^2{x} + x^2)^{\frac{1}{2}}\right]$

$\frac{1}{2}(\cot^2{x} + x^2)^{-\frac{1}{2}} \cdot (-2\cot{x}\csc^2{x} + 2x)$

I'll leave the very little bit of possible simplification to you.
• Sep 17th 2008, 05:48 PM
javax
$
(\sqrt{\cot^2 x+x^2}) = \frac{(\cot^2 x)'+(x^2)'}{2\sqrt{\cot^2 x+x^2}} = \frac{-2\cot x \csc^2 x + 2x}{2\sqrt{\cot^2 x+x^2}} = \frac{x-\cot x \csc^2 x}{\sqrt{\cot^2 x+x^2}}
$

I don't think it can get any simplier.
• Sep 17th 2008, 05:49 PM
javax
Quote:

Originally Posted by skeeter
do you know how the chain rule works?

$\frac{d}{dx} \left[(\cot^2{x} + x^2)^{\frac{1}{2}}\right]$

$\frac{1}{2}(\cot^2{x} + x^2)^{-\frac{1}{2}} \cdot (-2\cot{x}\csc^2{x} + 2x)$

I'll leave the very little bit of possible simplification to you.

sorry I didn't see your post when I started to post (Lipssealed)
• Sep 17th 2008, 05:50 PM
NotEinstein
Quote:

Originally Posted by skeeter
do you know how the chain rule works?

$\frac{d}{dx} \left[(\cot^2{x} + x^2)^{\frac{1}{2}}\right]$

$\frac{1}{2}(\cot^2{x} + x^2)^{-\frac{1}{2}} \cdot (-2\cot{x}\csc^2{x} + 2x)$

I'll leave the very little bit of possible simplification to you.

just learned it yesterday.. guess this is good practice. only question. why after you took the derivative of the outermost term, did you put * (-2cotx ...)? i understand that you brought the 2 to the front using the power rule.. but wouldn't it stay positive?

EDIT: Nevermind, i understand now. Thank you
• Sep 17th 2008, 05:56 PM
skeeter
chain rule inside the chain rule ...

the derivative of $\cot^2{x}$ is

$2\cot{x} \cdot (-csc^2{x}) = -2\cot^2{x}\csc^2{x}$
• Sep 17th 2008, 06:39 PM
john doe
I think you made a mistake you squared the cot