1. ## First Derivative

we have: $\displaystyle y= 10x ^{Ln10x}$
First thing we do: $Ln (y) = Ln(10x) * Ln(10x)$
Finally after applying the "rules" : $y' = y* [ \frac {2 Ln (10x)}{x} ]$

Am I correct? Because wolfram alpha gives the answer in a different form first derivative of y= 10x^(Ln10x) - Wolfram|Alpha

2. ## Re: First Derivative

Originally Posted by sakonpure6
we have: $\displaystyle y= 10x ^{Ln10x}$
First thing we do: $Ln (y) = Ln(10x) * Ln(10x)$
Finally after applying the "rules" : $y' = y* [ \frac {2 Ln (10x)}{x} ]$

Am I correct? Because wolfram alpha gives the answer in a different form first derivative of y= 10x^(Ln10x) - Wolfram|Alpha

$y=10 x^{\ln(10x)}=10 \exp(\ln(x)\ln(10x))$

$\dfrac {dy}{dx}=10\exp(\ln(x)\ln(10x))\times \dfrac {d}{dx}\left(\ln(x)\ln(10x)\right)=$

$10\exp(\ln(x)\ln(10x))\times\left(\dfrac 1 x \ln(10x)+\ln(x)\dfrac{10}{x}\right)=$

$10^{\ln(10x)} \left(\dfrac 1 x \ln(10x)+\ln(x)\dfrac{10}{x}\right)$

This is what Mathematica gives.. I 'm sure this an the various forms wolfram alpha gives are all equivalent.

3. ## Re: First Derivative

I presume that "log10x" means "log, base 10, of x" or $\displaystyle log_{10}(x)$. However, taking the logarithm of both sides does NOT give $\displaystyle log_{10}(x)\log_{10}(x)$. it gives, rather, $\displaystyle log_{10}(y)= log_{10}(10)+ log_{10}(x^{log_{10}(10x)= 1+ log_{10}(x)log_{10}(10x}= 1+ log_{10}(x)+ log_{10}(x)log_{10}(x)= 1+ log_{10}(x)+ (log_{10}(x))^2$.

The derivative of both sides of that, with respect to x, is $\displaystyle \frac{1}{y}\frac{dy}{dx}= \frac{ln(10)}{x}+ \frac{2ln(10)(log_{10}(x))}{x}$