Find frac{dy}{dx} if 2x^2 - 3xy + y^2 + x + 2 | vedclass

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

Question

(A) Given equation is $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \

Vedclass · Accepted Answer

$\frac{3y - 4x - 1}{2y - 3x + 2}$

Vedclass · Answer

(A) Given equation is $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0$ $4x - 3(y + x \frac{dy}{dx}) + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$ $4x - 3y - 3x \frac{dy}{dx} + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$ Grouping the $\frac{dy}{dx}$ terms: $(2y - 3x + 2) \frac{dy}{dx} = 3y - 4x - 1$ Therefore,$\frac{dy}{dx} = \frac{3y - 4x - 1}{2y - 3x + 2}$. Thus,option $A$ is correct.

Find $\frac{dy}{dx}$ if $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$.

Similar Questions

Find $\frac{dy}{dx}$ for the equation $x^{3}+x^{2}y+xy^{2}+y^{3}=81$.

If the tangent to the curve $xy + ax + by = 0$ at $(1, 1)$ makes an angle of $\tan^{-1} 2$ with the positive direction of the $x$-axis, then the value of $\frac{ab}{a + b}$ is...

If $\sqrt{y + x} + \sqrt{y - x} = c$, then $\frac{dy}{dx} = f(x) - \sqrt{[f(x)]^2 - 1}$. Find $f(x)$.

If $x=e^{(x/y)}$,then $\frac{dy}{dx}=$

The slope of the tangent drawn at $(1, 4)$ to the curve given implicitly by the equation $2(x \sqrt{y} + y \sqrt{x}) = 4x \sqrt{x} + y \sqrt{y}$ is -

Vedclass Test Series

Exam Paper Generator

Online Exam Module