Thank you for reporting, we will resolve it shortly
Q.
If $\sin x \cdot \cos y=\frac{1}{2}$, then the value of $\frac{ d ^2 y }{ dx ^2}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is equal to
Continuity and Differentiability
Solution:
$\cos x \cos y-\sin y \sin x \cdot \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\cot x \cdot \cot y$
$\text { and } \frac{ d ^2 y }{ dx ^2}=-\operatorname{cosec}^2 x \cot y -\operatorname{cosec}^2 y \cdot \cot x \cdot \frac{ dy }{ dx } $
$\because\left(\frac{ dy }{ dx }\right)_{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)}=1 $
$\therefore\left(\frac{ d ^2 y }{ dx ^2}\right)_{\left(\frac{\pi}{4}, \frac{\pi}{4}\right)}=-4 $