Question

If $K={\left(\sin y\right)}^{\sin \left(\frac{\pi }{2}x\right)}$, then at $x=1,\frac{dy}{dx}=$

• A. $p$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{2}\log K$
• D. $0$

Answer 1

The correct option is D $0$

$K=\left(\sin y\right)\sin \left(\frac{\pi x}{2}\right)$

$\log K=\sin \left(\frac{\pi x}{2}\right)\log \left(\sin y\right)$

Differentiate $\Rightarrow$

$0=\sin \left(\frac{\pi x}{2}\right)\frac{\cos y}{\sin y}\frac{dy}{dx}+\frac{\pi }{2}\cos \frac{\pi x}{2}\log \sin y$

$\Rightarrow \frac{dy}{dx}=\frac{-\frac{\pi }{2}\cos \left(\frac{\pi x}{2}\right)\log \left(\sin y\right)}{\sin \left(\frac{\pi x}{2}\right)\frac{\cos y}{\sin y}}$

at $x=1\Rightarrow K=\sin y$

$\therefore {\left(\frac{dy}{dx}\right)}_{x=1}=\frac{-\frac{\pi }{2}\cos \left(\frac{\pi }{2}\right)\log \left(K\right)}{\sin \left(\frac{\pi }{2}\right)\frac{\sqrt{1-K^2}}{K}}$

$=0$