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
$logx^y + logy^x = log2^{4}$
$\Rightarrow ylog\,x + xlog\,y = 4 \,log\,2$
Differentiating both sides $w$.$r$.$t$. $x$, we get
$\frac{y}{x}+log\,x \frac{dy}{dx}+log\,y\cdot1=0$
$\Rightarrow \frac{dy}{dx}=-\frac{\left(log\,y+\frac{y}{x}\right)}{\left(log\,x+\frac{x}{y}\right)}$
$\therefore \frac{dy}{dx}\bigg|_{(2,2)}$$=\frac{\left(log\,2+1\right)}{\left(log\,2+1\right)}=-1$