Continuity and Differentiability — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability2 Marks
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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