Q63 / 110 · VITEEE 2024 · Mathematics
If $(\cos x)^y=(\sin y)^x$, then $\frac{d y}{d x}$ equals
- A$\frac{\log \sin y-y \tan x}{\log \cos x+x \cot y}$
- Correct answerB$\frac{\log \sin y+y \tan x}{\log \cos x-x \cot y}$
- C$\frac{\log \sin y+y \tan x}{\log \cos x+x \cot y}$
- D$\frac{\log \sin y+y \tan x}{\log \cos y-y \cot x}$
Solution
$(\cos x)^y=(\sin y)^x$
Taking log on both sides
$y \ln (\cos x)=x \ln (\sin y)$
On differentiating both sides w.r.t $x$, we get
$\begin{aligned}
& y \cdot\left(\frac{1}{\cos x}(-\sin x)\right)+\frac{d y}{d x}(\ln (\cos x)) \\
& =x\left(\frac{1}{\sin y} \times \cos y\right) \frac{d y}{d x}+\ln (\sin y) \\
& \Rightarrow \frac{d y}{d x}\left(\ln (\cos x)-\frac{x \cos y}{\sin y}\right)=\ln (\sin y)+y \tan x \\
& \Rightarrow \frac{d y}{d x}(\ln (\cos x)-x \cot y)=\ln (\sin y)+y \tan x \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{\ln (\sin y)+y \tan x}{\ln (\cos x)-x \cot y}
\end{aligned}$
Hence, the correct answer is option: 'B'.