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?

  • $\begingroup$ 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

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 $(x\ln(2))'=\ln(2)$.

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

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 $\log_n{x}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.