On differentiation, if we obtain f(x, y) dy - | vedclass

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

Question

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

Vedclass · Accepted Answer

$-3$

Vedclass · Answer

(B) Given the equation: $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(2x^2) - \frac{d}{dx}(3xy) + \frac{d}{dx}(y^2) + \frac{d}{dx}(x) + \frac{d}{dx}(2y) - \frac{d}{dx}(8) = 0$. $4x - 3(x \frac{dy}{dx} + y) + 2y \frac{dy}{dx} + 1 + 2 \frac{dy}{dx} = 0$. Grouping the $\frac{dy}{dx}$ terms: $\frac{dy}{dx}(2y - 3x + 2) + (4x - 3y + 1) = 0$. This can be written as: $(2y - 3x + 2) dy + (4x - 3y + 1) dx = 0$. Comparing this with $f(x, y) dy - g(x, y) dx = 0$, we get: $f(x, y) = 2y - 3x + 2$ and $g(x, y) = -(4x - 3y + 1) = 3y - 4x - 1$. Now, calculate $g(2, 2) = 3(2) - 4(2) - 1 = 6 - 8 - 1 = -3$. Calculate $f(1, 1) = 2(1) - 3(1) + 2 = 2 - 3 + 2 = 1$. Therefore, $\frac{g(2, 2)}{f(1, 1)} = \frac{-3}{1} = -3$.

On differentiation, if we obtain $f(x, y) dy - g(x, y) dx = 0$ from $2x^2 - 3xy + y^2 + x + 2y - 8 = 0$, then find the value of $\frac{g(2, 2)}{f(1, 1)}$.

Similar Questions

If $\sin \left(\frac{x+y}{x-y}\right)=\tan \frac{\pi}{5}$,then $\frac{d y}{d x}=$

If ${x^m}{y^n} = {(x + y)^{m + n}}$,then ${\left. {\frac{dy}{dx}} \right|_{x = 1, y = 2}}$ is equal to:

If $xy \neq 0, x+y \neq 0$ and $x^m y^n=(x+y)^{m+n}$,where $m, n \notin N$,then $\frac{dy}{dx}$ is equal to

If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$ where $a>b>0$,then at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$,$\frac{dy}{dx}=$

The curve $y - e^{xy} + x = 0$ has a vertical tangent at

Vedclass Test Series

Exam Paper Generator

Online Exam Module