If e^{-y} cdot y = x,then frac{dy}{dx} is | vedclass

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

Question

(D) Given the equation $e^{-y} \cdot y = x$. Taking the derivative with respect to $x$ on both sides: $\frac{d}{dx}(y e^{-y}) = \frac{d}{dx}(x)$ Using

Vedclass · Accepted Answer

$\frac{y}{x(1-y)}$

Vedclass · Answer

(D) Given the equation $e^{-y} \cdot y = x$. Taking the derivative with respect to $x$ on both sides: $\frac{d}{dx}(y e^{-y}) = \frac{d}{dx}(x)$ Using the product rule: $y \cdot \frac{d}{dx}(e^{-y}) + e^{-y} \cdot \frac{dy}{dx} = 1$ $y \cdot (-e^{-y}) \frac{dy}{dx} + e^{-y} \frac{dy}{dx} = 1$ Factor out $\frac{dy}{dx}$: $\frac{dy}{dx} (e^{-y} - y e^{-y}) = 1$ $\frac{dy}{dx} e^{-y} (1 - y) = 1$ $\frac{dy}{dx} = \frac{1}{e^{-y}(1 - y)}$ Since $e^{-y} = \frac{x}{y}$,substitute this into the expression: $\frac{dy}{dx} = \frac{1}{(\frac{x}{y})(1 - y)} = \frac{y}{x(1 - y)}$.

If $e^{-y} \cdot y = x$,then $\frac{dy}{dx}$ is

Similar Questions

Consider the non-constant differentiable function $f$ of one variable which obeys the relation $\frac{f(x)}{f(y)}=f(x-y)$. If $f^{\prime}(0)=p$ and $f^{\prime}(5)=q$, then $f^{\prime}(-5)$ is

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

If $y = \operatorname{Sin}^{-1}\left(\frac{4+5 \sin x}{5+4 \sin x}\right)$,then find $\frac{dy}{dx}$.

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

If $x^{2}+y^{2}=1$,then $\frac{d^{2} x}{d y^{2}}=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module