1. ## d^2y/dx^2

find $\displaystyle d^2y/dx^2$ by implicit differentiation

$\displaystyle 3x^2-4y^2=7$

$\displaystyle 6x-8yy'=0$

$\displaystyle y'=\frac{6x}{8y}=-\frac{3x}{4y}$

$\displaystyle \frac{d^2y}{dx^2} = \frac{3x(4y')-4y(3)}{16y^2}$

i don't see this moving towards $\displaystyle -\frac{21}{16y^3}$which is the answer

2. In your formula for $\displaystyle \frac{d^2y}{dx^2}$, you have a y', which you can replace by what you found just above ! : $\displaystyle y'=-\frac{3x}{4y}$
And then finish it off with the original equation : $\displaystyle 3x^2-4y^2=7 \Rightarrow x^2=\frac{7+4y^2}{3}$

Plus, you applied the quotient rule incorrectly, the derivative of $\displaystyle \frac uv$ is $\displaystyle \frac{u'v-uv'}{v^2}$, but you did $\displaystyle uv'-u'v$

3. Plus, you applied the quotient rule incorrectly, the derivative of $\displaystyle \frac {u}{v}$ is $\displaystyle \frac{u'v-uv'}{v^2}$, but you did $\displaystyle uv'-u'v$
then
$\displaystyle \left(-\frac{3x}{4y}\right)' = -\frac{(3)4y-3x(4y')}{16y^2}$

4. Originally Posted by bigwave
then
$\displaystyle \left(-\frac{3x}{4y}\right)' = -\frac{(3)4y-3x(4y')}{16y^2}$
So sorry, there was indeed the minus sign...
Did you manage to go through all the calculations to get the desired answer ?

5. We have $\displaystyle 3x^2-4y^2=7$ and $\displaystyle y'=-\frac{3x}{4y}$

$\displaystyle \frac{d^2y}{dx^2}=\frac{3x(4y')-4y(3)}{16y^2}=\frac{12(xy'-y)}{16y^2}=\frac 34\cdot\frac{xy'-y}{y^2}$

But $\displaystyle y'=-\frac{3x}{4y}\Rightarrow xy'=-\frac{3x^2}{4y}\Rightarrow xy'-y=-\frac{3x^2}{4y}-y=-\frac{3x^2-4y^2}{4y}$

But the very first equation tells us that $\displaystyle 3x^2-4y^2=7$

Hence $\displaystyle xy'-y=-\frac{7}{4y}$

Therefore, $\displaystyle \frac{d^2y}{dx^2}=\frac 34\cdot\left(-\frac{7}{4y^3}\right)=-\frac{21}{16y^3}$