If ln left( {(e - 1){e^{xy}} + {x^2}} right) | vedclass

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

Question

(C) Given the equation: $\ln \left( {(e - 1){e^{xy}} + {x^2}} \right) = {x^2} + {y^2}$.First,we differentiate both sides with respect to $x$ using the

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) Given the equation: $\ln \left( {(e - 1){e^{xy}} + {x^2}} \right) = {x^2} + {y^2}$.First,we differentiate both sides with respect to $x$ using the chain rule:$\frac{1}{(e - 1){e^{xy}} + {x^2}} \cdot \left( {(e - 1){e^{xy}} \cdot \left( y + x \frac{dy}{dx} \right) + 2x} \right) = 2x + 2y \frac{dy}{dx}$.Now,substitute the point $(x, y) = (1, 0)$ into the equation:At $x = 1, y = 0$,the term inside the logarithm is $(e - 1){e^0} + 1^2 = e - 1 + 1 = e$.So,$\ln(e) = 1^2 + 0^2 = 1$,which is consistent.Substituting $(1, 0)$ into the differentiated equation:$\frac{1}{e} \cdot \left( {(e - 1){e^0} \cdot \left( 0 + 1 \cdot \frac{dy}{dx} \right) + 2(1)} \right) = 2(1) + 2(0) \cdot \frac{dy}{dx}$.$\frac{1}{e} \cdot \left( {(e - 1) \frac{dy}{dx} + 2} \right) = 2$.$(e - 1) \frac{dy}{dx} + 2 = 2e$.$(e - 1) \frac{dy}{dx} = 2e - 2$.$(e - 1) \frac{dy}{dx} = 2(e - 1)$.Therefore,$\left. \frac{dy}{dx} \right|_{(1,0)} = 2$.

If $\ln \left( {(e - 1){e^{xy}} + {x^2}} \right) = {x^2} + {y^2}$,then $\left. {\frac{{dy}}{{dx}}} \right|_{(1,0)}$ is

Similar Questions

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

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

If $\cos ^{-1}\left(\frac{x^2-y^2}{x^2+y^2}\right)=k$ (a constant), then $\frac{d y}{d x}$ is equal to

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

The curve $x^4-2xy^2+y^2+3x-3y=0$ cuts the $X$-axis at $(0,0)$ at an angle of

Vedclass Test Series

Exam Paper Generator

Online Exam Module