If e^x+e^y=e^{x+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: $e^x + e^y = e^{x+y}$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^

Vedclass · Accepted Answer

$-e^{y-x}$

Vedclass · Answer

(A) Given the equation: $e^x + e^y = e^{x+y}$ Differentiating both sides with respect to $x$: $\frac{d}{dx}(e^x) + \frac{d}{dx}(e^y) = \frac{d}{dx}(e^{x+y})$ $e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} \cdot (1 + \frac{dy}{dx})$ $e^x + e^y \cdot \frac{dy}{dx} = e^{x+y} + e^{x+y} \cdot \frac{dy}{dx}$ Rearranging the terms to isolate $\frac{dy}{dx}$: $e^y \cdot \frac{dy}{dx} - e^{x+y} \cdot \frac{dy}{dx} = e^{x+y} - e^x$ $\frac{dy}{dx} (e^y - e^{x+y}) = e^{x+y} - e^x$ Since $e^{x+y} = e^x + e^y$,substitute this into the equation: $\frac{dy}{dx} (e^y - (e^x + e^y)) = (e^x + e^y) - e^x$ $\frac{dy}{dx} (e^y - e^x - e^y) = e^y$ $\frac{dy}{dx} (-e^x) = e^y$ $\frac{dy}{dx} = -\frac{e^y}{e^x} = -e^{y-x}$

If $e^x+e^y=e^{x+y}$,then $\frac{dy}{dx}=$

Similar Questions

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

If $3^x y^x = x^{3y}$, then the value of $\frac{dy}{dx}$ at $x = 1$ is:

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

If $\log _{10}\left(\frac{x^2-y^2}{x^2+y^2}\right)=2$,then $\frac{dy}{dx} = \dots$

If $y^m + y^{-m} = 2x$, then $(x^2 - 1) \left( \frac{dy}{dx} \right)^2 = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module