Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Find derivative of $y=\ln(\tan^{-1}(2x^4))$

The answer is: $y^{\prime}=\dfrac{8x^3}{(4x^8+1)\tan^{-1}(2x^4)}$

Can you show the steps on how to get this answer?

I know that:

  • $\dfrac{d}{dx}(\tan^{-1}(x))=\dfrac{1}{1+x^2}$
  • $\dfrac{d}{dx}\ln(x)=\dfrac{y^{\prime}}{y}=\dfrac{1}{x}$

I think this problem requires use of the chain rule twice, so we have $y=f(u)=\ln(u), u=g(x)=\tan^{-1}(2x^4) \implies f(g(x)).$

Can you please show steps on how to solve this? Thank you.

share|cite|improve this question
up vote 1 down vote accepted

If you are new to chain rule, I would suggest you create new variables (in your head once you become good at it. So here they are $$ u = 2x^4\\ v= \tan^{-1} u \\ y=\ln v $$ to get $$ y = \ln(\tan^{-1}(2 x^4)$$

So chain rule says: $$ \frac{dy}{dx} = \frac{dy} {dv} \frac{dv}{du} \frac{du}{dx} = \frac{1}{v} \frac 1 {1+u^2} 8 x^3 $$ Now write everything in terms of $x$ to get your answer.

With practice you can skip introducing all these extra variables, but I would recommend you do this way till you gain confidence.

share|cite|improve this answer
+1 thanks, so you used the chain rule three times, correct? – Emi Matro Jan 2 '14 at 5:25
yes I did. Again, all these intermediate variables are not needed but they help you focus on the chain rule. – user44197 Jan 2 '14 at 5:32

You have all the steps, and just need to put it together:

\begin{align*} \frac{dy}{dx} &= \frac{1}{\tan^{-1}(2x^4)} \left(\frac{d}{dx} \tan^{-1}(2x^4)\right) \\ &=\frac{1}{\tan^{-1}(2x^4)} \frac{1}{1 + (2x^4)^2} \left(\frac{d}{dx} 2x^4\right) \\ &= \frac{1}{\tan^{-1}(2x^4)} \frac{1}{1 + (2x^4)^2} (8x^3) \end{align*}

share|cite|improve this answer
Thanks, helpful +1 – Emi Matro Jan 2 '14 at 5:58

You can also start taking the exponentials of the lhs and rhs. This will shorten the derivation steps and basically reduce to the first derivative you wrote.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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