For |x| < 1, the derivative of y = ln[ sqrt(1-x^2) ] is?

[Math wiz ya rite 09]

For |x| < 1, the derivative of y = ln[ sqrt(1-x^2) ] is?

 

[arthur ohlsten]

lxl<1 is the same as
-1<x<1

y=ln[1-x^2]^1/2 take derivative
dy/dx = [1/[1-x^2]^1/2] 1/2[1-x^2]^-1/2 [-2x]
dy/dx =[ 1/[1-x^2] ][-2x
dy/dx = -2x/[1-x^2] for -1<x<1 answer
Arthur

 

[soroban]

Re: derivative involving ln

Hello, Math wiz ya rite 09!

Differentiate: \(\displaystyle \:f(x) \;=\;\ln\left(\sqrt{1\,-\,x^2}\right)\)

We have: \(\displaystyle \:f(x) \;=\;\ln (1\,-\,x^2 )^{\frac{1}{2}} \;=\;\frac{1}{2}\cdot\ln (1\,-\,x^2 )\)

Chain Rule: \(\displaystyle \;f'(x)\;=\;\frac{1}{2}\,\cdot\,\frac{1}{1\,-\,x^2}\,\cdot\,(-2x) \;=\;\frac{-x}{1\,-\,x^2}\)

Therefore: \(\displaystyle \L\:f'(x) \;=\;\frac{x}{x^2\,-\,1}\)

 

[Math wiz ya rite 09]

didnt you forget the sqrt when converting the ln to a fraction?

 

[arthur ohlsten]

not to interfere , but he did.

Simple error
Arthur

 

[skeeter]

no, he didn't forget the square root.

he changed \(\displaystyle \L \ln{\sqrt{1-x^2}}\) to \(\displaystyle \L \frac{1}{2} \ln(1-x^2)\) using the power law for logs.

the derivative is correct.

 

[arthur ohlsten]

you are correct. I didn't follow his math, to see he rewrote the equation. I guess because I didn't.
His approach is more elegant than mine. Sorry for the comment

Arthur