Thread: Differentiate

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

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

Originally Posted by Punch

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

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

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

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

Originally Posted by BAdhi

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

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

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

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

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

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

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

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

Have I got it right of worse?

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

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

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

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

thanks.

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