If x^y cdot y^x = 16,then frac{dy}{dx} at (2, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Given the equation $x^y \cdot y^x = 16$. Taking the natural logarithm on both sides,we get: $y \ln x + x \ln y = \ln 16$. Differentiating both sid

Vedclass · Accepted Answer

-$1$

Vedclass · Answer

(A) Given the equation $x^y \cdot y^x = 16$. Taking the natural logarithm on both sides,we get: $y \ln x + x \ln y = \ln 16$. Differentiating both sides with respect to $x$ using the product rule: $\frac{d}{dx}(y \ln x) + \frac{d}{dx}(x \ln y) = \frac{d}{dx}(\ln 16)$. $\left( \frac{dy}{dx} \ln x + y \cdot \frac{1}{x} \right) + \left( 1 \cdot \ln y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} \right) = 0$. Rearranging the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} (\ln x + \frac{x}{y}) = -(\frac{y}{x} + \ln y)$. $\frac{dy}{dx} = -\frac{\frac{y}{x} + \ln y}{\ln x + \frac{x}{y}}$. Now,substitute the point $(2, 2)$ into the expression: $\left( \frac{dy}{dx} \right)_{(2, 2)} = -\frac{\frac{2}{2} + \ln 2}{\ln 2 + \frac{2}{2}} = -\frac{1 + \ln 2}{1 + \ln 2} = -1$.

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

Similar Questions

If ${x^m}{y^n} = {(x + y)^{m + n}}$,then ${\left. {\frac{dy}{dx}} \right|_{x = 1, y = 2}}$ is equal to:

If $2x^2 - 3xy + 4y^2 + 2x - 3y + 4 = 0$,then $\left(\frac{dy}{dx}\right)_{(3,2)} = $

If $y \sec x + \tan x + x^2 y = 0$,then $\frac{dy}{dx} =$

If $x^2 \tan ^{-1} \frac{y}{x}-y^2 \tan ^{-1} \frac{x}{y}=k$,then $\left(\frac{d y}{d x}\right)_{(1,1)}=$

If $x^{2}+y^{2}=t+\frac{1}{t}$ and $x^{4}+y^{4}=t^{2}+\frac{1}{t^{2}}$,then find $\frac{d y}{d x}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module