clovnerie0q

2022-10-07

Use Implicit differentiation to prove
${y}^{″}=\frac{-9}{{y}^{2}}$ if $4{x}^{2}-2{y}^{2}=9$

Kaitlyn Levine

differentiate both sides of: $4{x}^{2}-2{y}^{2}=9$
$\frac{d}{dx}\left(4{x}^{2}-2{y}^{2}=9\right)\phantom{\rule{0ex}{0ex}}8x-4y{y}^{\prime }=0$
If you don't understand why $\frac{d}{dx}{y}^{2}=2y{y}^{\prime }$ then you need to talk to your teacher / TA / tutor to explain the concept of implicit differentiation again. If that is the case don't stress, it is a little slippery.
isolate ${y}^{\prime }$
${y}^{\prime }=\frac{2x}{y}$
Now differentiate again (with respect to x) to find ${y}^{″}$
${y}^{″}=\frac{2y-2x{y}^{\prime }}{{y}^{2}}$
and substitute the value of ${y}^{\prime }$ found above.
${y}^{″}=\frac{2y-2x\frac{2x}{y}}{{y}^{2}}\phantom{\rule{0ex}{0ex}}{y}^{″}=\frac{2{y}^{2}-4{x}^{2}}{{y}^{3}}$
and $2{y}^{2}-4{x}^{2}=-9$ from the original constraint.
${y}^{″}=\frac{-9}{{y}^{3}}$

Sara Solomon

assuming that
$y=y\left(x\right)$
is given then we get
$8x-4y{y}^{\prime }=0$

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