x^3+y^3=3xy Rightarrow frac{dy}{dx}= | vedclass

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

Question

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

Vedclass · Accepted Answer

$\frac{y-x^2}{y^2-x}$

Vedclass · Answer

(A) Given the equation: $x^3+y^3=3xy$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(3xy)$. $3x^2 + 3y^2 \frac{dy}{dx} = 3(y + x \frac{dy}{dx})$. Dividing by $3$: $x^2 + y^2 \frac{dy}{dx} = y + x \frac{dy}{dx}$. Rearranging the terms to isolate $\frac{dy}{dx}$: $y^2 \frac{dy}{dx} - x \frac{dy}{dx} = y - x^2$. $\frac{dy}{dx}(y^2 - x) = y - x^2$. Therefore,$\frac{dy}{dx} = \frac{y-x^2}{y^2-x}$.

$x^3+y^3=3xy \Rightarrow \frac{dy}{dx}=$

Similar Questions

$f(x)$ and $g(x)$ are two differentiable functions on $[0, 2]$ such that $f''(x) - g''(x) = 0$,$f'(1) = 2$,$g'(1) = 4$,$f(2) = 3$,and $g(2) = 9$. Then $f(x) - g(x)$ at $x = 3/2$ is:

If $y = \sqrt[3]{\tan x + y}$, then $\frac{dy}{dx} =$

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

If ${2^x} + {2^y} = {2^{x + y}},$ then the value of $\frac{dy}{dx}$ at $x = y = 1$ is

If $x+y=\tan ^{-1} y$ and $\frac{d^{2} y}{d x^{2}}=f(y) \frac{d y}{d x}$,then $f(y)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module