Question

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

• A. –1
• B. 0
• C. 1
• D. None of these

Answer 1

Answer: (A)

–1

Answer 2

x^y y^x = 16

∴ log_e x^y + log_e y^x = log_e 16

⇒ y log_e x + x log_e y = 4 log_e 2

Now differentiating both side w.r.t.x

$\frac{y}{x}$ + log_e x $\frac{dy}{dx}$ + $\frac { x } { y }$ $\frac{dy}{dx}$ + log_e y . 1 = 0

∴ $\frac{dy}{dx}$ = – $\frac{\left( \log_{e}y + \frac{y}{x} \right)}{\left( \log_{e}x + \frac{x}{y} \right)}$

∴ $\left. \ \frac{dy}{dx} \right|_{(2,2)}$= – $\frac{(\log_{e}2 + 1)}{(\log_{e}2 + 1)}$ = – 1.