If e^y + xy = e, then the ordered pair (frac{ | vedclass

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

Question

(B) Step $1$: Find $y$ at $x = 0$. Substituting $x = 0$ into $e^y + xy = e$, we get $e^y + 0 = e$, so $y = 1$.Step $2$: Differentiate $e^y + xy = e$ w

Vedclass · Accepted Answer

$(\frac{-1}{e}, \frac{1}{e^2})$

Vedclass · Answer

(B) Step $1$: Find $y$ at $x = 0$. Substituting $x = 0$ into $e^y + xy = e$, we get $e^y + 0 = e$, so $y = 1$.Step $2$: Differentiate $e^y + xy = e$ with respect to $x$: $e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$.Step $3$: At $x = 0, y = 1$, substitute these into the derivative: $e^1 \frac{dy}{dx} + 1 + 0 = 0 \implies \frac{dy}{dx} = -\frac{1}{e}$.Step $4$: Differentiate $e^y \frac{dy}{dx} + y + x \frac{dy}{dx} = 0$ again: $e^y (\frac{dy}{dx})^2 + e^y \frac{d^2y}{dx^2} + \frac{dy}{dx} + \frac{dy}{dx} + x \frac{d^2y}{dx^2} = 0$.Step $5$: Substitute $x = 0, y = 1, \frac{dy}{dx} = -\frac{1}{e}$ into the second derivative equation: $e^1(-\frac{1}{e})^2 + e^1 \frac{d^2y}{dx^2} + 2(-\frac{1}{e}) + 0 = 0$.Step $6$: Simplify: $\frac{1}{e} + e \frac{d^2y}{dx^2} - \frac{2}{e} = 0 \implies e \frac{d^2y}{dx^2} = \frac{1}{e} \implies \frac{d^2y}{dx^2} = \frac{1}{e^2}$.Step $7$: The ordered pair is $(-\frac{1}{e}, \frac{1}{e^2})$.

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

Similar Questions

If $x = e^{\tan^{-1}\left(\frac{y-x^2}{x^2}\right)}$,then $\frac{dy}{dx}$ at $x = 1$ is

The two curves $x^{3}-3xy^{2}+2=0$ and $3x^{2}y-y^{3}=2$:

If $\sin (xy) + \frac{x}{y} = {x^2} - y,$ then $\frac{dy}{dx} = $

If the tangent to the curve $xy + ax + by = 0$ at $(1, 1)$ makes an angle $\tan^{-1}(2)$ with the $X$-axis,then $\frac{ab}{a+b} =$ ?

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module