# Thread: derivative problem with ln

1. ## derivative problem with ln

Differentiate. .

$1/(xsqrt(x^2-2))(sqrt(x^2-2) + (2x^2)/2sqrt(x^2-2)$

common denominator: $2sqrt(x^2-2)$
= $1/(xsqrt(x^2-2)) {[2(x^2-2)+x^2 ]/[2sqrt(x^2-2)}$

getting rid of the other denominator
= $3x^2-4/(2*sqrt(x^2-2))*(1/1/(xsqrt(x^2-2))$
= $3x^2-4/ 2x(x^2-2)$
= $3x^2-4/2x^3-4x$

2. Originally Posted by hazecraze
Differentiate. .

$1/(xsqrt(x^2-2))(sqrt(x^2-2) + (2x^2)/2sqrt(x^2-2)$

common denominator: $2sqrt(x^2-2)$
= $1/(xsqrt(x^2-2)) {[2(x^2-2)+x^2 ]/[2sqrt(x^2-2)}$

You seem to have an error in here when you calculated the derivative of $\sqrt{x^2-2}$. It seems like the 2 should have been canceled.

getting rid of the other denominator
= $3x^2-4/(2*sqrt(x^2-2))*(1/1/(xsqrt(x^2-2))$
= $3x^2-4/ 2x(x^2-2)$
= $3x^2-4/2x^3-4x$
$g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+x\frac{d}{dx}\sqrt{x^2-2})$

$\frac{d}{dx}\sqrt{x^2-2}=\frac{2x}{2\sqrt{x^2-2}}=\frac{x}{\sqrt{x^2-2}}$

So this becomes:

$g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+\frac{x^2}{\sqrt{x^2-2}})$

$g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+x\frac{d}{dx}\sqrt{x^2-2})$

$\frac{d}{dx}\sqrt{x^2-2}=\frac{2x}{2\sqrt{x^2-2}}=\frac{x}{\sqrt{x^2-2}}$

So this becomes:

$g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+\frac{x^2}{\sqrt{x^2-2}})$
Ohh, so... $g'(x)=\frac{2x^2-2}{2\sqrt{x^2-2}}$
cancel out the 2s

$g'(x)=\frac{x^2-1}{\sqrt{x^2-2}}$

=

$=\frac{x^2-1}{\sqrt{x^2-2}}$* $=\frac{1}{x\sqrt{x^2-2}}$

= $g'(x)=\frac{x^2-1}{x^3-2}$

4. Originally Posted by hazecraze
Ohh, so... $g'(x)=\frac{2x^2-2}{2\sqrt{x^2-2}}$
cancel out the 2s

$g'(x)=\frac{x^2-1}{\sqrt{x^2-2}}$

=

$=\frac{x^2-1}{\sqrt{x^2-2}}$* $=\frac{1}{x\sqrt{x^2-2}}$

= $g'(x)=\frac{x^2-1}{x^3-2}$
You simplified this wrong:

$g'(x)=\frac{1}{x\sqrt{x^2+2}}(\frac{2x^2-2}{\sqrt{x^2-2}})$
$g'(x)=\frac{2x^2-2}{x(x^2-2)}$