Question

If $xy^2=1$, prove that $2\dfrac{dy}{dx}+y^3=0$​

Answer 1

$xy^2=1---(1)$

$\implies y^2=\dfrac{1}{x}----(2)$

Differentiate eq(1) with respect to x :

$\dfrac{d}{dx}(xy^2)=\dfrac{d}{dx}(1)\\\\\implies x(\dfrac{d}{dx}(y^2))+y^2(\dfrac{d}{dx}(x))=0\\\\\implies 2xy\dfrac{dy}{dx}+y^2=0\\\\\implies y(2x(\dfrac{dy}{dx})+y)=0\\\\\implies 2x(\dfrac{dy}{dx})+y=0\\\\\textbf{From eq(2) we get :}\\\\\implies \dfrac{2}{y^2}(\dfrac{dy}{dx})+y=0\\\\\implies 2\dfrac{dy}{dx}+y^3=0$

NOTE :

$\dfrac{d}{dx}(1)=0$

$\dfrac {d}{dx}}(u\cdot v)={\dfrac {du}{dx}}\cdot v+u\cdot {\dfrac {dv}{dx}$

Answer 2

Explanation:

Given xy² = 1

We have to find the derivative of it.

On differentiating both sides, we get

$\Longrightarrow \frac{d}{dx}(xy^2) = \frac{d}{dx}(1)$

$\Longrightarrow \frac{d}{dx}(x)y^2 + \frac{d}{dx}(y^2)x = 0$

$\Longrightarrow 1.y^2 + 2y \frac{d}{dx}(y)x = 0$

$\Longrightarrow y^2 + 2xy \frac{d}{dx}(y) = 0$

$\Longrightarrow y^2 + 2xyy' = 0$

$\Longrightarrow y^2 = -2xyy'$

$\Longrightarrow y' = -\frac{y}{2x}$

$\Longrightarrow \frac{d}{dx}(y) = -\frac{y}{2x}$

LHS:

$\Longrightarrow 2\frac{dy}{dx} + y^3$

$\Longrightarrow 2(-\frac{y}{2x}) + y^3$

$\Longrightarrow -\frac{y}{x} + y^3$

$\Longrightarrow y(y^2 - \frac{1}{x})$

$\Longrightarrow y(\frac{1}{x} - \frac{1}{x}) \ [ \because xy^2 = 1 \rightarrow y^2 = \frac{1}{x}]$

$\Longrightarrow \textbf {0}$

RHS

Hope it helps!