Question

If $x^y\cdot y^x=16$, then the value of $\frac{dy}{dx}$ at $\left(2,2\right)$ is

• A. $-1$
• B. $0$
• C. $1$
• D. None of these

Answer 1

Answer: (A)

Since, $x^y\cdot y^x=16$
Taking log on both sides, we get
$log{x}^y+log{y}^x=log{2}^4$
$\Rightarrow ylogx+xlogy=4log2$
Differentiating both sides $w$.$r$.$t$. $x$, we get
$\frac{y}{x}+logx\frac{dy}{dx}+logy\cdot 1=0$
$\Rightarrow \frac{dy}{dx}=-\frac{\left(logy+\frac{y}{x}\right)}{\left(logx+\frac{x}{y}\right)}$
$\therefore \frac{dy}{dx}{|}_{(2,2)}$$=\frac{\left(log2+1\right)}{\left(log2+1\right)}=-1$