Differentiate

**Punch** · Jun 12th 2011, 12:05 AM

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

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

**Amer** · Jun 12th 2011, 01:02 AM

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
$\frac{dy}{dx}=\left(\frac{lnx}{ln(x-1)}\right)^{'}= \frac{(ln(x-1))(\frac{1}{x})-lnx(\frac{1}{x-1})}{[ln(x-1)]^2}$

**BAdhi** · Jun 12th 2011, 01:07 AM

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

**Prove It** · Jun 12th 2011, 03:30 AM

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.

**BAdhi** · Jun 12th 2011, 04:39 AM

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

**Prove It** · Jun 12th 2011, 04:46 AM

$\displaystyle \ln{(x)}$ 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 $\displaystyle \ln{|x|}$ ?

**BAdhi** · Jun 12th 2011, 05:22 AM

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?

**Prove It** · Jun 12th 2011, 05:33 AM

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...

**BAdhi** · Jun 12th 2011, 05:54 AM

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.

**Prove It** · Jun 12th 2011, 06:12 AM

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

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