Derivative of the function $\sin^{-1}\Big(\frac{2^{x+1}}{1+4^x}\Big)$

Question

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 ?

Answer 1

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

Answer 2

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

Answer 3

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

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 ;

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