Question

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

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

Answer 1

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

Given,
$y^x=x^{\sin y}$
Taking $\ln$ on both sides, we get
$x\ln y=\sin y\ln x$.

Differentiate w.r.t. $x$
$\ln y+x\cdot \frac{1}{y}\cdot \frac{dy}{dx}=\sin y\cdot \frac{1}{x}+\ln x\cdot \cos y\cdot \frac{dy}{dx}$

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

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