Tough Derivative

Calculus Level 2

If $x^{2} + y^{2} = 4$ , then find $y'$ .

y' = \frac{y}{x}

y' = \frac{x}{y}

y'= -\frac{y}{x}

y' = -\frac{x}{y}

2 solutions

Colin Carmody
Mar 9, 2016

$x^{2} + y^{2} = 4$

Then take the derivative of both sides.

$2x\frac{dx}{dx} + 2y\frac{dy}{dx} = 0$

$x + yy^{,} = 0$

$yy^{,} = -x$

$y^{,} = \boxed{-\frac{x}{y}}$

Arkajyoti Banerjee
May 3, 2016

We are given, $\displaystyle x^{2} + y^{2} = 4$ .

Differentiating both sides with respect to $x$ , we have:

$\implies \displaystyle \frac{d}{dx}x^{2} + \frac{d}{dx}y^{2} = \frac{d}{dx}4x^{0}$

$\implies \displaystyle 2x + 2y\frac{dy}{dx} = 0$

$\implies \displaystyle x + y\frac{dy}{dx} = 0$

$\implies \displaystyle \frac{dy}{dx} = -\frac{x}{y}$

$\implies \displaystyle \boxed{y' = -\frac{x}{y}}$