1. Differentiate

Differentiate $y=\frac{lnx}{ln(x-1)}$ with respects to x

$y=\frac{lnx}{ln(x-1)}= \frac{(ln(x-1))(\frac{1}{x})-lnx(\frac{1}{x-1})}{[ln(x-1)]^2}$

2. Originally Posted by Punch
Differentiate $y=\frac{lnx}{ln(x-1)}$ with respects to x

$y=\frac{lnx}{ln(x-1)}= \frac{(ln(x-1))(\frac{1}{x})-lnx(\frac{1}{x-1})}{[ln(x-1)]^2}$
it is correct but make it well order
$\frac{dy}{dx}=\left(\frac{lnx}{ln(x-1)}\right)^{'}= \frac{(ln(x-1))(\frac{1}{x})-lnx(\frac{1}{x-1})}{[ln(x-1)]^2}$

3. also it has to be $\ln{|x-1|}$ not $\ln{(x-1)}$

4. Originally Posted by BAdhi
also it has to be $\ln{|x-1|}$ not $\ln{(x-1)}$
No it doesn't. Only if you are arriving at a logarithm through integration do you need the modulus signs.

Only if you are arriving at a logarithm through integration do you need the modulus signs.
Prove It, can you please explain it more because I quite don't get it

6. $\displaystyle \ln{(x)}$ is a continuous function on $\displaystyle (0, \infty)$ and is differentiable on $\displaystyle (0, \infty)$. Its derivative is $\displaystyle \frac{1}{x}$.

But $\displaystyle \frac{1}{x}$ is defined on all $\displaystyle x \in \mathbf{R}\backslash \{0\}$, so should have an antiderivative on $\displaystyle (-\infty, 0)$ as well.

So we need to think outside the box a little. Remember that $\displaystyle |x| = \begin{cases} x \textrm{if }x \geq 0\\ -x \textrm{if }x < 0\end{cases}$.

What is the derivative of $\displaystyle \ln{|x|}$?

7. It should be $\frac{1}{|x|}$.

Ok, let me get this straight. When you mention $\ln{(x)}$, variable x's domain should always be $(0,\infty)$. And the derivative of it then will become $\frac{1}{x}$. For a variable $y \in (-\infty,\infty)$ it should be $\ln{|y|}$ and the derivative is $\frac{1}{|y|}$ (we can verify it by knowing the slope of the curve $\ln{(x)}$ is always positive).

Have I got it right of worse?

8. No, the derivative is not $\displaystyle \frac{1}{|x|}$.

Remember that $\displaystyle \ln{|x|} = \ln{(x)}$ if $\displaystyle x > 0$ and $\displaystyle \ln{|x|} = \ln{(-x)}$ if $\displaystyle x < 0$.

So to evaluate the derivative of $\displaystyle \ln{|x|}$, you need to differentiate each of those two cases. You should notice something...

9. sorry sir, forgot that the curve of the $\ln{|x|}$ is $\ln{(x)}$ mirrored around the y axis. So derivative should still be $\frac{1}{x}$

thanks.

10. Correct, so the antiderivative of $\displaystyle \frac{1}{x}$ is $\displaystyle \ln{|x|}$