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How to compute $f'(x)$ where $f(x)=\log_{x}(x^2+3)$ ?

When we deal with $x^{x^x}$ we use $e^{x^x\ln x}$. What do we "do" with logarithms?

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Try using a change of base formula ($\log_b a={\ln a\over \ln b}$, e.g.). –  David Mitra Jan 18 '13 at 16:39
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$f(x)=\log_u(u^2+3)$ has derivative $f'(x)=0$... :) –  AndreasT Jan 18 '13 at 16:41
    
@AndreasT Why is that? From what David Mitra suggested I got $f'(x)= \frac{ \frac {2xlnx}{x^2+3} - \frac{ln(x^2+3)}{x}}{ln^2x}$ $= \frac{2u^2lnu-(u^2+3)ln(u^2+3)}{ln^2u}$ –  Hagrid Jan 18 '13 at 17:01
    
@Hagrid I know, I was just joking about your mistyped variable $u$, for which $f$ would be constant. Your question has already been edited though... –  AndreasT Jan 18 '13 at 17:04
    
Very nice :) Are my calculations in the comment above correct? –  Hagrid Jan 18 '13 at 17:05
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2 Answers 2

up vote 4 down vote accepted

Use the change-of-base formula: $$ \log_x (x^2+3) = \frac{\log_e (x^2+3)}{\log_e x}\quad\left(\text{or, if you like, }\frac{\ln(x^2+3)}{\ln x}\right). $$

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Let $y=f(x)=\log_{x}(x^2+3)\implies x^y=x^2+3$. Taking derivative on both sides gives, $$x^y(\frac{y}{x}+y'\log x)=2x\implies (x^2+3)(\frac{y}{x}+y'\log x)=2x$$ Now you can solve for $y'$ easily.

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