clovnerie0q

2022-10-07

Use Implicit differentiation to prove

${y}^{\u2033}=\frac{-9}{{y}^{2}}$ if $4{x}^{2}-2{y}^{2}=9$

${y}^{\u2033}=\frac{-9}{{y}^{2}}$ if $4{x}^{2}-2{y}^{2}=9$

Kaitlyn Levine

Beginner2022-10-08Added 12 answers

differentiate both sides of: $4{x}^{2}-2{y}^{2}=9$

$\frac{d}{dx}(4{x}^{2}-2{y}^{2}=9)\phantom{\rule{0ex}{0ex}}8x-4y{y}^{\prime}=0$

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 ${y}^{\u2033}$

${y}^{\u2033}=\frac{2y-2x{y}^{\prime}}{{y}^{2}}$

and substitute the value of ${y}^{\prime}$ found above.

${y}^{\u2033}=\frac{2y-2x\frac{2x}{y}}{{y}^{2}}\phantom{\rule{0ex}{0ex}}{y}^{\u2033}=\frac{2{y}^{2}-4{x}^{2}}{{y}^{3}}$

and $2{y}^{2}-4{x}^{2}=-9$ from the original constraint.

${y}^{\u2033}=\frac{-9}{{y}^{3}}$

$\frac{d}{dx}(4{x}^{2}-2{y}^{2}=9)\phantom{\rule{0ex}{0ex}}8x-4y{y}^{\prime}=0$

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 ${y}^{\u2033}$

${y}^{\u2033}=\frac{2y-2x{y}^{\prime}}{{y}^{2}}$

and substitute the value of ${y}^{\prime}$ found above.

${y}^{\u2033}=\frac{2y-2x\frac{2x}{y}}{{y}^{2}}\phantom{\rule{0ex}{0ex}}{y}^{\u2033}=\frac{2{y}^{2}-4{x}^{2}}{{y}^{3}}$

and $2{y}^{2}-4{x}^{2}=-9$ from the original constraint.

${y}^{\u2033}=\frac{-9}{{y}^{3}}$

Sara Solomon

Beginner2022-10-09Added 1 answers

assuming that

$y=y(x)$

is given then we get

$8x-4y{y}^{\prime}=0$

$y=y(x)$

is given then we get

$8x-4y{y}^{\prime}=0$

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