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I have a couple of small questions regarding differentiating logarithmic functions:

  • The derivative of $ \log(x)^2 = \dfrac{2}{\ln(10)x}$

  • The derivative of $ 2 \log(x) = \dfrac{2}{\ln(10)x}$

Does this hold for any $n$?

This is a problem I found in my textbook, I have to differentiate the following function:

$f(x) = \ln(2^x)$

The answer is :

$[f(x) = \ln(2^x)]' = x.\ln(2) = \ln(2)$

I really don't understand the first step, can anyone explain this and give the intuition behind this answer?

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  • $\begingroup$ What is $n$? And I take it that , by $\log{x}$, you mean $\log_{10}{x}$. $\endgroup$ – Ron Gordon Jan 15 '13 at 14:07
  • $\begingroup$ When you write $\log(x)^2$, do you mean $(\log x)^2$ or $\log(x^2)$? It's horribly ambiguous. $\endgroup$ – Michael Hardy Jan 15 '13 at 15:12
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You noted that:

  • The derivative of $\log(x)^2 = \dfrac{2}{\ln(10)x}$

I think this is not right unless there is a typo in it. In fact we have $$\left((\log(x))^2\right)'=2\times\log(x)\times\frac{1}{x\ln(10)}$$

About your question, note that $\ln(a^b)=b\ln(a), a>0$, so $\ln(2^x)=x\ln(2)$. $\ln(2)$ is just a constant, so as we know $(ax)'=a$, then $(x\ln(2))'=\ln(2)$.

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  • $\begingroup$ well done...and almost at 14k! $\endgroup$ – Namaste Feb 17 '13 at 0:04
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Use the fact that

$$\log{x} = \frac{\ln{x}}{\ln{10}} $$

and use the chain rule:

$$ \frac{d}{dx} \log^2{x} = \frac{1}{\ln^2{10}} (2 \ln{x}) \frac{1}{x} $$

You can generalize the first equation for any base of $\log_n{x}$.

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