smuklica8i

2022-07-22

use implicit differentiation to find $\frac{dy}{dx}$ in terms of $x$ and $y$

1.) ${x}^{3}-xy+{y}^{2}=4$

2.) $y=\mathrm{sin}\left(xy\right)$
find $\frac{{d}^{2}y}{d{x}^{2}}$

3.) ${x}^{2}{y}^{2}-2x=3$

Jazlene Dickson

Expert

I'll show you how to do 1), and you can try the others yourself. Differentiating implicitly, we have
$\frac{d}{dx}\left({x}^{3}-xy+{y}^{2}\right)=\frac{d}{dx}4$
or
$3{x}^{2}-y-x{y}^{\prime }+2y{y}^{\prime }=0$
so that
$3{x}^{2}-y={y}^{\prime }\left(x-2y\right)$
and hence
${y}^{\prime }=\frac{3{x}^{2}-y}{x-2y}.$

Livia Cardenas

Expert

And 2,
$dy=\mathrm{cos}\left(xy\right)\ast \left(ydx+xdy\right)=y\mathrm{cos}\left(xy\right)dx+x\mathrm{cos}\left(xy\right)dy$
using chain and product rules. Then solve for $dy$, and then finally $\frac{dy}{dx}$,
$dy-x\mathrm{cos}\left(xy\right)dy=\mathrm{cos}\left(xy\right)\ast \left(ydx+xdy\right)=y\mathrm{cos}\left(xy\right)dx⇒dy\left(1-x\mathrm{cos}\left(xy\right)\right)=y\mathrm{cos}\left(xy\right)dx⇒\frac{dy}{dx}=\frac{y\mathrm{cos}\left(xy\right)}{1-x\mathrm{cos}\left(xy\right)}$

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