1. derivative check

$\displaystyle y = \frac{e^xlogx}{log(2x+1)}$

$\displaystyle y '= \frac{e^x(logx+\frac{1}{x})log(2x+1)-(\frac{2}{2x+1})(e^xlogx)}{(log(2x+1))^2}$

can i then cancel out the log(2x+1) from the top and one from the bottom leaving the denominator as log(2x+1) ?? thank u

2. This is a much easier way...I don't think you took your derivative right because the derivative of logx is not 1/x. It would be right if you change your logs to ln.
$\displaystyle y = \frac{e^xlogx}{log(2x+1)}$

$\displaystyle y = \frac{\frac{e^xlnx}{ln10}}{\frac{ln(2x+1)}{ln10}}$

$\displaystyle y = \frac{e^xlnx}{ln(2x+1)}$

$\displaystyle y '= \frac{e^x(lnx+\frac{1}{x})ln(2x+1)-(\frac{2}{2x+1})(e^xlnx)}{(ln(2x+1))^2}$

I don't see anything you can cancel out.

3. Originally Posted by Linnus
This is a much easier way...I don't think you took your derivative right because the derivative of logx is not 1/x. It would be right if you change your logs to ln.
$\displaystyle y = \frac{e^xlogx}{log(2x+1)}$

$\displaystyle y = \frac{\frac{e^xlnx}{ln10}}{\frac{ln(2x+1)}{ln10}}$

$\displaystyle y = \frac{e^xlnx}{ln(2x+1)}$

$\displaystyle y '= \frac{e^x(lnx+\frac{1}{x})ln(2x+1)-(\frac{2}{2x+1})(e^xlnx)}{(ln(2x+1))^2}$$\displaystyle$

I don't see anything you can cancel out.
so the ln(2x+1) dont cancel out ?

on this website (solving derivatives step-by-step) says the differentiation of logx is 1/x ? even on my answer paper for this question it says 1/x

4. Originally Posted by jvignacio
so the ln(2x+1) dont cancel out ?

on this website (solving derivatives step-by-step) says the differentiation of logx is 1/x ? even on my answer paper for this question it says 1/x
$\displaystyle \frac {dy}{dx} lnx=\frac {1}{x}$
$\displaystyle \frac {dy}{dx} log_bx=\frac {1}{x}lnb$
No it doesn't cancel out because you cannot factor out a ln(2x+1) on the top.

It says on the website "On this site log[x] is the same as ln[x]"

5. Originally Posted by jvignacio
so the ln(2x+1) dont cancel out ?

on this website (solving derivatives step-by-step) says the differentiation of logx is 1/x ? even on my answer paper for this question it says 1/x
If $\displaystyle \log$ represents $\displaystyle \log_e$ then the derivative of $\displaystyle \log x$ is $\displaystyle \frac{1}{x}$.

Note: $\displaystyle \log_e$ is usually represented by $\displaystyle \ln$. Hence the comments previously posted.

To answer your question, consider ..... Can you take $\displaystyle \log (2x + 1)$ out as a common factor in the numerator? No? Then you can't cancel it out.

6. none

$\displaystyle \frac {dy}{dx} log_bx=\frac {1}{x}lnb$
This seems a little bit misleading but maybe I am reading it poorly.
dy/dx(logx) is actually 1/x * 1/lnx, not 1/x * lnx which would put the lnx in the numerator instead of the denominator.

7. Originally Posted by gml
$\displaystyle \frac {dy}{dx} log_bx=\frac {1}{x}lnb$
This seems a little bit misleading but maybe I am reading it poorly.
dy/dx(logx) is actually 1/x * 1/lnx, not 1/x * lnx which would put the lnx in the numerator instead of the denominator.
Did you mean $\displaystyle \frac{d \log x}{dx}= (1/x) (1/\ln(10))$? It is certainly NOT $\displaystyle (1/x) (1/\ln(x))$.

If $\displaystyle y= \log(x)$ then $\displaystyle x= 10^y= e^{\ln(10^y)}= e^{yln(10)}$ so, by the chain rule, $\displaystyle \frac{dx}{dy}= \ln(10) e^{y \ln (10)}= \ln(10) 10^y= \ln (10) x$. Thus, if $\displaystyle y= \log (x)$, $\displaystyle \frac{dy}{dx}= \frac{1}{x \ln(10)}$.