Problem:
$\displaystyle $y=ln(x\sqrt{x^{2}-1}$)$
I can get this far:
$\displaystyle $y=ln(x)+ln(x^{2}-1)^{1/2}$$
$\displaystyle $=ln(x)+(1/2)ln(x^{2}-1)$$
But I get confused on where to go from here. Thanks in advance for any help.
Problem:
$\displaystyle $y=ln(x\sqrt{x^{2}-1}$)$
I can get this far:
$\displaystyle $y=ln(x)+ln(x^{2}-1)^{1/2}$$
$\displaystyle $=ln(x)+(1/2)ln(x^{2}-1)$$
But I get confused on where to go from here. Thanks in advance for any help.
Sorry, should have mentioned it was a differentiation problem. I get this far:
$\displaystyle $y=ln(x\sqrt{x^{2}-1}$)$
$\displaystyle $y=ln(x)+ln(x^{2}-1)^{1/2}$$
$\displaystyle $=ln(x)+(1/2)ln(x^{2}-1)$$
$\displaystyle $y'=1/x+((1/2)/(x^{2}-1))$ $
Is this right so far?
I'm not asking for tutoring. If I was ready to do this question I wouldn't have had to post it.
I figured it out eventually though. Thanks for the help guys.
$\displaystyle y=\frac{1}{2}ln(x^{2}-1)$
$\displaystyle y'=(0)(ln(x^{2}-1)+(\frac{1}{2})(\frac{2x}{x^{2}-2})$
$\displaystyle =\frac{2x}{2x^{2}-2}$
$\displaystyle =\frac{2(x)}{2(x^{2}-1)}$
$\displaystyle =\frac{x}{x^{2}-1}$
The purpose of logarithmic differentiation is to make use of logarithm rules to simplify the function, so that difficult functions can be differentiated more easily.
Make use of the fact that $\displaystyle \displaystyle \ln{(x^2 - 1)} = \ln{[(x - 1)(x + 1)]} = \ln{(x- 1)} + \ln{(x + 1)}$.
I'm sure you can differentiate the last term...