Limit Continuity And Differentiability Ques 22
- Find $\frac{d y}{d x}$ at $x=-1$, when
$ (\sin y)^{\sin \frac{\pi}{2} x}+\frac{\sqrt{3}}{2} \sec ^{-1}(2 x)+2^x \tan \ln (x+2)=0 . $
(1991, 4M)
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Answer:
Correct Answer: 22.$(\frac{3}{\pi \sqrt{\pi^2-3}})$
Solution: Here, $(\sin y)^{\sin \frac{\pi}{2} x}+\frac{\sqrt{3}}{2} \sec ^{-1}(2 x)+2^x \tan \{\log (x+2)\}=0$
On differentiating both sides, we get
$(\sin y)^{\sin \frac{\pi}{2} x} \cdot \log (\sin y) \cdot \cos \frac{\pi}{2} x \cdot \frac{\pi}{2}$ $+\left(\sin \frac{\pi}{2} x\right)(\sin y)^{\left(\sin \frac{x}{2} x\right)-1} \cdot \cos y \cdot \frac{d y}{d x} $
$+\frac{\sqrt{3}}{2} \cdot \frac{2}{(2|x|) \sqrt{4 x^2-1}}+\frac{2^x \cdot \sec ^2\{\log (x+2)\}}{(x+2)} $ $+2^x \log 2 \cdot \tan \{\log (x+2)\}=0$
Putting $\left(x=-1, y=-\frac{\sqrt{3}}{\pi}\right)$, we get
$ \frac{d y}{d x}=\frac{\left(-\frac{\sqrt{3}}{\pi}\right)^2}{\sqrt{1-\left(\frac{\sqrt{3}}{\pi}\right)^2}}=\frac{3}{\pi \sqrt{\pi^2-3}} $
BETA
