Question

Find $\frac{dy}{dx}$ at $x=-1$, when
$(\sin y{)}^{\sin \frac{\pi }{2}x}+\frac{\sqrt{3}}{2}{\sec }^{-1}(2x)+2^x\tan \ln (x+2)=0$.

Answer 1

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