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

Let $y=y(x)$ be a function of $x$ satisfying $y \sqrt{1-x^{2}}=k-x \sqrt{1-y^{2}}$ where $k$ is a constant and $y(\frac{1}{2})=-\frac{1}{4}.$ Then $\frac{dy}{dx}$ at $x=\frac{1}{2}$ is equal to:

If $y = x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \frac{1}{x^2 + \dots \infty}}}$,then $\frac{dy}{dx} = $

The focal length of a mirror is given by $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$. In finding the values of $u$ and $v$, the errors are equal to $p$. Then, the relative error in $f$ is

If $\sin(x + y) + \cos(2x + 2y) = \ln(3x + 3y)$,then $\frac{dy}{dx}$ is

$x\sqrt{1 + y} + y\sqrt{1 + x} = 0$,then $\frac{dy}{dx} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module