How do you find $\frac{d y}{d x}$ $dy/dx$ by implicit differentiation of $x^{2} + y^{2} = 36$ $x^2+y^2=36$ ?

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Answer 1 · 2017-01-20T06:12:50

$\frac{d y}{d x} = - \frac{y}{x}$ $dy/dx = -y/x$

$\frac{d}{d x} (x^{2} + y^{2}) = \frac{d}{d x} 36$ $d/dx(x^2+y^2) = d/dx36$

$\Rightarrow \frac{d}{d x} x^{2} + \frac{d}{d x} y^{2} = 0$ $=> d/dxx^2 + d/dxy^2 = 0$

$\Rightarrow 2 x + 2 y (\frac{d}{d x} y) = 0$ $=> 2x + 2y(d/dxy) = 0$
(Applying the chain rule with $y$ $y$ treated as a function of $x$ $x$ )

$\Rightarrow 2 x + 2 y \frac{d y}{d x} = 0$ $=> 2x+2ydy/dx = 0$

$\Rightarrow 2 y \frac{d y}{d x} = - 2 x$ $=> 2ydy/dx = -2x$

$:. dy/dx = -x/y$

How do you find dy/dxdydxdy/dx by implicit differentiation of x^2+y^2=36x2+y2=36x^2+y^2=36?