Find frac{dy}{dx} for the function y^{x} = x^ | vedclass

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

Question

(A) Given the function $y^{x} = x^{y}$.Taking the natural logarithm on both sides,we get:$x \log y = y \log x$Differentiating both sides with respect

Vedclass · Accepted Answer

$\frac{y}{x} \left( \frac{y - x \log y}{x - y \log x} \right)$

Vedclass · Answer

(A) Given the function $y^{x} = x^{y}$.Taking the natural logarithm on both sides,we get:$x \log y = y \log x$Differentiating both sides with respect to $x$ using the product rule:$\log y \cdot \frac{d}{dx}(x) + x \cdot \frac{d}{dx}(\log y) = \log x \cdot \frac{d}{dx}(y) + y \cdot \frac{d}{dx}(\log x)$$\log y + x \cdot \frac{1}{y} \cdot \frac{dy}{dx} = \log x \cdot \frac{dy}{dx} + y \cdot \frac{1}{x}$Rearranging the terms to isolate $\frac{dy}{dx}$:$\frac{x}{y} \frac{dy}{dx} - \log x \frac{dy}{dx} = \frac{y}{x} - \log y$$\left( \frac{x}{y} - \log x \right) \frac{dy}{dx} = \frac{y}{x} - \log y$$\left( \frac{x - y \log x}{y} \right) \frac{dy}{dx} = \frac{y - x \log y}{x}$Therefore,$\frac{dy}{dx} = \frac{y}{x} \left( \frac{y - x \log y}{x - y \log x} \right)$.

Find $\frac{dy}{dx}$ for the function $y^{x} = x^{y}$.

Similar Questions

If $2 x^2-3 x y+y^2+x+2 y-8=0$,then $\frac{d y}{d x}$ is equal to

If $e^{y} + xy = e$,then the ordered pair $\left(\frac{dy}{dx}, \frac{d^2y}{dx^2}\right)$ at $x = 0$ is equal to

$If \frac{x}{x-y} = \log \left(\frac{a}{x-y}\right)$,then $\frac{dy}{dx} =$

If $\cos (x + y) = y\sin x,$ then $\frac{dy}{dx} = $

If $y^{\cos x}=x^{\sin y}$, then $\frac{d y}{d x}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module