Question

find derivative dy/dx if x^(3)-3xy+y^(3)=2

Answer 1

$\bf x^{3} - 3xy + {y}^{3} = 2 \\ \\ {\sf { \underline{ \red{Differentiate \: w.r.t \: \: x}}}} \\ \\ \bf \frac{d {x}^{3} }{dx}+ \dfrac{3dxy}{dx} + \dfrac{dy^{3} }{dx} = 0 \\ \\ {\sf { \underline{ \red{Applying \: product \: rule}}}} \\ \\ \bf 3{x}^{2} +3 \[ \left [ x \dfrac{dy}{dx} + y \dfrac{dx}{dx} \right] \] + 3 {y}^{2} = 0 \\ \\ \bf 3{x}^{2} +3 \[ \left [ x \dfrac{dy}{dx} + y \right] \] + 3 {y}^{2} = 0 \\ \\ \bf 3{x}^{2} +3x \dfrac{dy}{dx} + 3y + 3 {y}^{2} = 0 \\ \\ \bf 3x \dfrac{dy}{dx} = - 3{x}^{2} - 3y - 3 {y}^{2} \\ \\ \bf \dfrac{dy}{dx} = \dfrac{3(-{x}^{2} -y - {y}^{2})}{3x} \\ \\ \bf \dfrac{dy}{dx} = \dfrac{(-{x}^{2} -y - {y}^{2})}{x} \\ \\ \bf \dfrac{dy}{dx} = \dfrac{(-{x}^{2}) }{x} - \dfrac{y}{x} - \dfrac{ {y}^{2} }{x} \\ \\ \bf \dfrac{dy}{dx} = - x - \dfrac{y}{x} - \dfrac{ {y}^{2} }{x}$

• Quick review

°•Above question is in the parametric form

°•Differentiate both sides i. e LHS and RHS

°•Differentiation of xⁿ = nx^n-1

°•Product rule = uv' + vu'