Answer
$\displaystyle \frac{dy}{dx}=x$
Work Step by Step
Differentiating an equation containing
terms of form $f(x)$ and $g(y)$,
terms $f(x)$ are differentiated directly, $\displaystyle \frac{d}{dx}[f(x)],$
and
terms $g(y)$ are differentiated using the chain rule,
$\displaystyle \frac{d}{dx}[g(y)]=\frac{d}{dy}[g(y)]\cdot\frac{dy}{dx}$
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$\displaystyle \frac{d}{dx}(x^{2}-2y)=\frac{d}{dx}(6)$
$2x-2\displaystyle \frac{dy}{dx}=0\qquad /-2x$
$-2\displaystyle \frac{dy}{dx}=-2x\qquad/\div(-2)$
$\displaystyle \frac{dy}{dx}=x.$
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Alternatively, solving the initial equation for y:
$x^{2}-2y=6\quad/-x^{2}$
$-2y=-x^{2}+6\quad/\div(-2)$
$y=\displaystyle \frac{1}{2}x^{2}-3$
Differentiating (constant m. power, constant),
$\displaystyle \frac{dy}{dx}=\frac{1}{2}\cdot 2x+0$
$\displaystyle \frac{dy}{dx}=x$
