Question

If $y^x=x^{\sin y}$, find $\frac{dy}{dx}$.

• A. $\frac{y}{x}\left[\frac{x\log y-\sin y}{y\log x\cos y-x}\right]$
• B. $\frac{y}{x}\left[\frac{x\log y+\sin y}{y\log x\cos y+x}\right]$
• C. $\frac{-y}{x}\left[\frac{x\log y-\sin y}{y\log x\cos y-x}\right]$
• D. $\frac{y}{x}\left[\frac{x\log y-\sin y}{y\log x\cos y+x}\right]$

Answer 1

The correct option is A $\frac{y}{x}\left[\frac{x\log y-\sin y}{y\log x\cos y-x}\right]$

Given $y^x=x^{\sin y}$.

Take log both sides.

$x\log y=\sin y\log x.$

Differentiate w.r.t. x

$\log y+x.\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}\sin y+\left(\log x\right)\cos y.\frac{dy}{dx}$

$(\log y-\frac{\sin y}{x})=\frac{dy}{dx}[\cos y\log x-\frac{x}{y}]$

$\therefore \frac{dy}{dx}=\frac{y}{x}\left[\frac{x\log y-\sin y}{y\log x\cos y-x}\right]$