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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.

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3 Answers 3

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.

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+1 thanks, so you used the chain rule three times, correct? –  user437158 Jan 2 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 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*}

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Thanks, helpful +1 –  user437158 Jan 2 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.

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