Question

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 $.

Answer 1

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
$(\sin y)^{\sin \frac{\pi}{2} x} \cdot \log (\sin y) \cdot \cos \frac{\pi}{2} x \cdot \frac{\pi}{2}$
$+\left(\sin \frac{\pi}{2} x\right)(\sin y)^{\left(\sin \frac{\pi}{2} x\right)-1} \cdot \cos y \cdot \frac{d y}{d x}$
$+\frac{\sqrt{3}}{2} \cdot \frac{2}{(2|x|) \sqrt{4 x^{2}-1}}+\frac{2^{x} \cdot \sec ^{2}\{\log (x+2)\}}{(x+2)}$
$+2^{x} \log 2 \cdot \tan \{\log (x+2)\}=0$
Putting $\left(x=-1, y=-\frac{\sqrt{3}}{\pi}\right)$, we get
$\frac{d y}{d x}=\frac{\left(-\frac{\sqrt{3}}{\pi}\right)^{2}}{\sqrt{1-\left(\frac{\sqrt{3}}{\pi}\right)^{2}}}=\frac{3}{\pi \sqrt{\pi^{2}-3}}$