Question

If $y^{\cos x}=x^{\sin y}$, then $\frac{dy}{dx}=$

• A. $\frac{y(x\sin x\log y+\sin y)}{x(\cos x-y\log x\cos y)}$
• B. $\frac{y(x\sin x\log x-\sin y)}{x(\cos x+y\log x\cos y)}$
• C. $\frac{y(\sin y-x\log y)}{x(x-y\cos y(\log x))}$
• D. $\frac{y(\sin y+x\log y)}{x(x+y\cos y(\log x))}$

Answer 1

Answer: (A)

It is given that
$y^{\cos x}=x^{\sin y}$
On taking logrithm both sides, we get
$\cos x\log y=\sin y\log x$
On differentiating both side w.r.t. $x$, we get
$\frac{1}{y}(\cos x)\frac{dy}{dx}-(\sin x)\log y$
$=\frac{1}{x}\sin y+(\log x)(\cos y)\frac{d}{dx}$
$\Rightarrow \frac{dy}{dx}\left(\frac{1}{y}\cos x-(\log x)\cos y\right)=\frac{1}{x}\sin y+(\sin x)\log y$
$\Rightarrow \frac{dy}{dx}=\frac{y(\sin y+x\sin x\log y)}{x(\cos x-y\log x\cos y)}$