Search any question & find its solution
Question:
Answered & Verified by Expert
If $y^{\cos x}=x^{\sin y}$, then $\frac{d y}{d x}=$
Options:
Solution:
2064 Upvotes
Verified Answer
The correct answer is:
$\frac{y(x \sin x \log y+\sin y)}{x(\cos x-y \log x \cos y)}$
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 $\frac{1}{y}(\cos x) \frac{d y}{d x}-(\sin x) \log y$
$=\frac{1}{x} \sin y+(\log x)(\cos y) \frac{d}{d x}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}\left(\frac{1}{y} \cos x-(\log x) \cos y\right)=\frac{1}{x} \sin y+(\sin x) \log y \\ & \Rightarrow \frac{d y}{d x}=\frac{y(\sin y+x \sin x \log y)}{x(\cos x-y \log x \cos y)}\end{aligned}$
$y^{\cos x}=x^{\sin y}$
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 $\frac{1}{y}(\cos x) \frac{d y}{d x}-(\sin x) \log y$
$=\frac{1}{x} \sin y+(\log x)(\cos y) \frac{d}{d x}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}\left(\frac{1}{y} \cos x-(\log x) \cos y\right)=\frac{1}{x} \sin y+(\sin x) \log y \\ & \Rightarrow \frac{d y}{d x}=\frac{y(\sin y+x \sin x \log y)}{x(\cos x-y \log x \cos y)}\end{aligned}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.