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

Question

Having trouble with this calculus question if someone could walk it through step by step I'd greatly appreciate it.

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

Answer 1

assuming that $$y=y(x)$$ is given then we get $$8x-4yy'=0$$

Answer 2

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

$\frac d{dx}(4x^2 - 2y^2 = 9)\\ 8x - 4yy' = 0$

If you don't understand why $\frac {d}{dx} y^2 = 2yy'$ 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 $y'$

$y' = \frac {2x}{y}$

Now differentiate again (with respect to x) to find $y''$

$y'' = \frac {2y - 2xy'}{y^2}$

and substitute the value of $y'$ found above.

$y'' = \frac {2y - 2x\frac {2x}{y}}{y^2}\\ y'' = \frac {2y^2 - 4x^2}{y^3}$

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

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

And your proposition is not true!