
derivative logarithm
Hi guys, I'm stuck :(
can someone please help me understand this:
f(x) = e^(2x) log(x^2)
we are given the answer which should be : f(x)' = ( e^(2x) / ln(10) ) * (ln(x^2) + 1)
this is where I'm stuck (I'm not even sure this is right), after the product rule:
f(x)' = 2e^(2x) * ( ln(x^2)) / ln(10) ) + ( (e^(2x) * 2x) / (x^2 ln(10)) )
I am on the right track here? Thanks

Re: derivative logarithm
$\displaystyle f(x) = e^{2x} \cdot \log{x^2}$
$\displaystyle f(x) = e^{2x} \cdot 2\log{x}$
log is base 10, change base ...
$\displaystyle f(x) = e^{2x} \cdot \frac{2}{\ln(10)} \cdot \ln{x}$
$\displaystyle f(x) = \frac{2}{\ln(10)}\left(e^{2x} \cdot \ln{x}\right)$
$\displaystyle f'(x) = \frac{2}{\ln(10)}\left(e^{2x} \cdot \frac{1}{x} + 2e^{2x} \cdot \ln{x}\right)$
$\displaystyle f'(x) = \frac{2e^{2x}}{\ln(10)}\left(\frac{1}{x} + 2\ln{x}\right)$

Re: derivative logarithm
Hi,
my first question:
What is the basis fo logaritm, 10 or e(Euler number) or what?
Than is the derivate of f(x) with the product und chain Regula to solve.
When lg=log with basis 10 and ln=log with basis e, we have:
lg(x)= lg(e)*ln(x)
When log = log of basis b, we have
log(x)= log(10)*lg(x)= log(e)*ln(x)
For derivating transform log in ln.
(e^2x)'= 2*e^2x
( ln(x))' = 1/x, where x >0
Don't help it?

Re: derivative logarithm
Thank you very much kind sir! This really helps to finally see what I was doing wrong. Much appreciated!