Question

If $x^y.y^x=16$ then $\frac{dy}{dx}$ at (2, 2) is

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

Answer 1

Answer: (B)

$-1$

$x^y.y^x=16$
$\log x^y+\log y^x=\log 16$
$y\log x+x\log y=\log 16$
Differentiating w.r.t x , we get
$\frac{y}{x}+\log x\frac{dy}{dx}+\frac{x}{y}\frac{dy}{dx}+\log y=0$
So, at x=2, y=2
$1+\log 2{\left(\frac{dy}{dx}\right)}_{(2,2)}+1{\left(\frac{dy}{dx}\right)}_{(2,2)}+\log 2=0$
$\Rightarrow {\left(\frac{dy}{dx}\right)}_{(2,2)}=-1$