derivative problem with ln

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• Oct 19th 2009, 02:15 PM
hazecraze
derivative problem with ln
Differentiate. http://www.webassign.net/cgi-bin/sym...%20-%202%29%29.

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

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

getting rid of the other denominator
=$\displaystyle 3x^2-4/(2*sqrt(x^2-2))*(1/1/(xsqrt(x^2-2))$
=$\displaystyle 3x^2-4/ 2x(x^2-2)$
=$\displaystyle 3x^2-4/2x^3-4x$
• Oct 19th 2009, 02:45 PM
adkinsjr
Quote:

Originally Posted by hazecraze
Differentiate. http://www.webassign.net/cgi-bin/sym...%20-%202%29%29.

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

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

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

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

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

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

So this becomes:

$\displaystyle g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+\frac{x^2}{\sqrt{x^2-2}})$
• Oct 19th 2009, 03:26 PM
hazecraze
Quote:

Originally Posted by adkinsjr
$\displaystyle g'(x)=\frac{1}{x\sqrt{x^2+2}}(\sqrt{x^2-1}+x\frac{d}{dx}\sqrt{x^2-2})$

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

So this becomes:

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

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

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

=

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

= $\displaystyle g'(x)=\frac{x^2-1}{x^3-2}$
• Oct 19th 2009, 03:59 PM
adkinsjr
Quote:

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

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

=

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

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

You simplified this wrong:

http://www.mathhelpforum.com/math-he...58d99b74-1.gif

Adding inside the parenthesis:

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

Now just multiply the fractions:

(this is where you messed up I think. You added the denominators instead of multiplying.)

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