If (x - y)e^{x/(x - y)} = k,then: | vedclass

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

Question

(B) Given the equation: $(x - y)e^{x/(x - y)} = k$. Taking the natural logarithm on both sides: $\ln(x - y) + \frac{x}{x - y} = \ln k$. Differentiatin

Vedclass · Accepted Answer

$y\frac{dy}{dx} + x - 2y = 0$

Vedclass · Answer

(B) Given the equation: $(x - y)e^{x/(x - y)} = k$. Taking the natural logarithm on both sides: $\ln(x - y) + \frac{x}{x - y} = \ln k$. Differentiating both sides with respect to $x$: $\frac{1}{x - y}(1 - \frac{dy}{dx}) + \frac{(x - y)(1) - x(1 - \frac{dy}{dx})}{(x - y)^2} = 0$. Multiply the entire equation by $(x - y)^2$: $(x - y)(1 - \frac{dy}{dx}) + (x - y) - x(1 - \frac{dy}{dx}) = 0$. $(x - y) - (x - y)\frac{dy}{dx} + x - y - x + x\frac{dy}{dx} = 0$. Combine the terms involving $\frac{dy}{dx}$: $(-x + y + x)\frac{dy}{dx} + (x - y + x - y - x) = 0$. $y\frac{dy}{dx} + x - 2y = 0$.

If $(x - y)e^{x/(x - y)} = k$,then:

Similar Questions

If $y=1+xe^y$,then $\frac{dy}{dx}=$

The equation of the normal to the curve $y=(1+x)^{2y}+\cos^{2}(\sin^{-1} x)$ at $x=0$ is

If $x{e^{xy}} = y + {\sin ^2}x$,then at $x = 0$,$\frac{dy}{dx} = $

For the curve $C : (x^{2}+y^{2}-3)+(x^{2}-y^{2}-1)^{5}=0$,the value of $3y^{\prime}-y^{3}y^{\prime\prime}$ at the point $(\alpha, \alpha)$,where $\alpha > 0$,on $C$ is equal to:

If $x^2+y^2=t+\frac{1}{t}$ and $x^4+y^4=t^2+\frac{1}{t^2}$,then $\frac{dy}{dx}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module