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Find y' if $y=\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$

In my reference $y'$ is given as $\frac{2^{x+1}\log2}{(1+4^x)}$. But is it a complete solution ?

Attempt 1

Let $2^x=\tan\alpha$ $$ \begin{align} y=\sin^{-1}\Big(\frac{2\tan\alpha}{1+\tan^2\alpha}\Big)=\sin^{-1}(\sin2\alpha)&\implies \sin y=\sin2\alpha=\sin\big(2\tan^{-1}2^x\big)\\ &\implies y=n\pi+(-1)^n(2\tan^{-1}2^x) \end{align} $$ $$ \begin{align} y'&=\pm\frac{2.2^x.\log2}{1+4^x}=\pm\frac{2^{x+1}.\log2}{1+4^x}\\ &=\color{blue}{\begin{cases} \frac{2^{x+1}.\log2}{1+4^x}\text{ if }-n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq -n\pi+\frac{\pi}{2}\\ -\frac{2^{x+1}.\log2}{1+4^x}\text{ if }n\pi-\frac{\pi}{2}\leq2\tan^{-1}2^x\leq n\pi+\frac{\pi}{2} \end{cases}} \end{align} $$ Attempt 2

$$ \begin{align} y'&=\frac{1}{\sqrt{1-\frac{(2^{x+1})^2}{(1+4^x)^2}}}.\frac{d}{dx}\frac{2^{x+1}}{1+4^x}\\ &=\frac{1+4^x}{\sqrt{1+4^{2x}+2.4^x-4^x.4}}.\frac{(1+4^x)\frac{d}{dx}2^{x+1}-2^{x+1}\frac{d}{dx}(1+4^x)}{(1+4^x)^2}\\ &=\frac{(1+4^x).2^{x+1}.\log2-2^{x+1}.4^x.\log2.2}{\sqrt{1+4^{2x}-2.4^x}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1+4^x-2.4^x\big]}{\sqrt{(1-4^x)^2}.(1+4^x)}\\&=\frac{2^{x+1}\log2\big[1-4^x\big]}{|{(1-4^x)}|.(1+4^x)}\\ &=\color{blue}{\begin{cases}\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1>4^x>0\\ -\frac{2^{x+1}\log2}{(1+4^x)}\text{ if }1<4^x \end{cases}} \end{align} $$ In both my attempts i am getting both +ve and -ve solutions. Is it the right way to find the derivative?

And how do I connect the domain for each cases in attempt 1 and attempt 2 ?

Sooraj S
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  • Holy cow. Is this an exercise from a book? Anyway, you may also try to take the sine on both sides and then use implicit differentiation. The neat thing is that $2^{x+1}$ and $1+4^x$ relate very nice in a right triangle. That is, through the Pythagorean theorem, the adjacent turns out to be $1-4^x$ (a little algebra) – imranfat Mar 30 '18 at 02:40
  • @imranfat could u pls elaborate a bit on how to follow this ?.. Could u pls comment on the domain of the $y'$ in both my attempts ?.. and is there both +ve and -ve solutions to the problem ? – Sooraj S Mar 30 '18 at 03:30
  • I can't find error...I checked it by desmos. Actually it is normal for analysis to have $\vert$(somethings)$\vert$. – Tony Ma Mar 30 '18 at 03:44
  • @imranfat thanx. so u mean final answer in both my attempts are correct right ? – Sooraj S Mar 30 '18 at 05:19
  • See https://math.stackexchange.com/questions/1764431/solving-arcsin-left2x-sqrt1-x2-right-2-arcsin-x and https://math.stackexchange.com/questions/979149/tan-frac12-sin-1-2x-1-x2-frac12-cos-1-1-y2-1 – lab bhattacharjee Mar 30 '18 at 05:43
  • @labbhattacharjee thnks for the help. could u pls confirm final answer in both my attempts are correct ? – Sooraj S Mar 30 '18 at 12:57

3 Answers3

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The trickiest part is the ^2 part let me know if you do not follow that step. I do not know how to use the fancy formatting so I had to write it out.

Solution

user9501082
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  • thanx. but how can it be different from my Attempt 2 ? – Sooraj S Mar 30 '18 at 03:17
  • I believe you simplified the inverse of ONLY Sin wrong. You can't bring in other items into the square root that easy without simplifying other things. Everything on the right side is correct, the left side is incorrect. – user9501082 Mar 30 '18 at 03:25
  • srry i could not follow u. could u pls mention which step seems wrong ?. are u saying the final answer $\pm\frac{2^{x+1}.\log2}{1+4^x}$ is wrong ? – Sooraj S Mar 30 '18 at 03:31
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$$y=\sin^{-1} \Big(\frac{2^{x+1}}{1+4^x}\Big)$$ the right side of the equation will be called $b$ it will help later. Take sine of both sides and the derivative $$\cos(y) y' =\frac{ln(2)2^{x+1}}{4^x+1}-\frac{ln(4) 4^x 2^{x+1}}{(4^x+1)^2}$$ the right side of the equation will be called $b'$ so $$ y' cos(y)=b'$$ divide both sides by cosine $$y'=\frac{b'}{\cos(y)}$$ substitute $y$ for $b$ and you have $$y'=\frac{b'}{cos(b)}$$

Latin Wolf
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$y = arcsin(\frac{2^{x+1}}{1+4^x})\\\implies y = arcsin (\frac{2.2^x}{1+(2^x)^2})$

consider $y = arcsin(\frac{2x}{1+x^2})$

here let $x = tan(\theta) \implies \theta = arctan(x)$

$y= arcsin(\frac{2tan(\theta)}{1+ tan^2(\theta)}) = arcsin(sin(2\theta)) = 2\theta = 2arctan(x)$

so in your given question, $y = arcsin(\frac{2^{x+1}}{1+4^x}) = 2arctan(2^x) $

differentiate wrt x ;

$y' = \frac{2^{x+1}ln(2)}{1+4^x}$

The Integrator
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  • $\sin^{-1}(\sin 2\theta)=2\theta$ only if $2\theta\in[-\pi/2,\pi/2]$ so i think doing that is just incomplete. – Sooraj S Mar 30 '18 at 09:10