If y=1+xe^y,then frac{dy}{dx}= | vedclass

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

Question

(A) Given the equation: $y = 1 + xe^y$ Differentiating both sides with respect to $x$ using the product rule: $\frac{dy}{dx} = 0 + \left( x \cdot e^y

Vedclass · Accepted Answer

$\frac{e^y}{2-y}$

Vedclass · Answer

(A) Given the equation: $y = 1 + xe^y$ Differentiating both sides with respect to $x$ using the product rule: $\frac{dy}{dx} = 0 + \left( x \cdot e^y \frac{dy}{dx} + e^y \cdot 1 \right)$ $\frac{dy}{dx} = xe^y \frac{dy}{dx} + e^y$ Rearranging the terms to isolate $\frac{dy}{dx}$: $\frac{dy}{dx} - xe^y \frac{dy}{dx} = e^y$ $\frac{dy}{dx} (1 - xe^y) = e^y$ $\frac{dy}{dx} = \frac{e^y}{1 - xe^y}$ From the original equation,we know $xe^y = y - 1$. Substituting this into the expression: $\frac{dy}{dx} = \frac{e^y}{1 - (y - 1)}$ $\frac{dy}{dx} = \frac{e^y}{1 - y + 1}$ $\frac{dy}{dx} = \frac{e^y}{2 - y}$

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

Similar Questions

For $x \in R$,$f(x) = |\log 2 - \sin x|$ and $g(x) = f(f(x))$,then

If $a(4+x^2)=x$ and $y-x^3=a^2$,then $\frac{dy}{dx}$ at $x=1$ is ...

If $x^{\frac{2}{5}}+y^{\frac{2}{5}}=a^{\frac{2}{5}}$,then $\frac{dy}{dx} = $

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

If $a f(x)+b f\left(\frac{1}{x}\right)=x+1$, and $\frac{d}{d x}\left(x^2 f(x)\right)=2 x^2+2 x+\frac{1}{3}$, then $a-b=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module