Differentiate the sin^ –1 ( 2^ x+1 1+4^ x ) w.r.t. x. - Vidyadip

Question

Differentiate the $sin^{–1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$ w.r.t. x.

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Answer

Let $f(x) = \sin^{–1}\left(\frac{2^{x+1}}{1+4^{x}}\right)$. To find the domain of this function we need to find all x such that $-1 \leq \frac{2^{x+1}}{1+4^{x}} \leq 1$. Since the quantity in the middle is always positive,
We need to find all x such that $\frac{2^{x+1}}{1+4^{x}} \leq$ 1, i.e., all x such that $2^{x + 1} \leq 1 + 4x$. We may rewrite this as 2 $\leq \frac{1}{2^{x}}$ + $2^x$ which is true for all x. Hence the function is defined at every real number. By putting 2x = tan $\theta$, this function may be rewritten as
$f(x)=\sin ^{-1}\left[\frac{2^{x+1}}{1+4^{x}}\right]$
= $\sin ^{-1}\left[\frac{2^{x} \cdot 2}{1+\left(2^{x}\right)^{2}}\right]$
= $\sin ^{-1}\left[\frac{2 \tan \theta}{1+\tan ^{2} \theta}\right]$
= $\sin^{–1} [sin 2\theta$]
= $2\theta = 2 \tan^{–1} (2^x)$
Thus $f^{\prime}(x)=2 \cdot \frac{1}{1+\left(2^{x}\right)^{2}} \cdot \frac{d}{d x}\left(2^{x}\right)$
= $\frac{2}{1+4^{x}} \cdot\left(2^{x}\right) \log 2$
= $\frac{2^{x+1} \log 2}{1+4^{x}}$

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