Finding the second derivative of a function (implicit differentiation) I am not sure if I got the correct answer or not. It was a homework from my textbook, and it does not have answers for even number questions...
Original Function: $x^2y-4x=5$ 
$\frac{dy}{dx}$ = $\frac{4-2xy}{x^2}$ 
It seems I got to finding the dy/dx part correctly according to wolfram alpha 
Heres the steps to finding $\frac{dy^2}{d^2x}$ 


*

*$\frac{(x^2)(-2y-2x\frac{dy}{dx}) - ((-2xy+4)(2x))}{x^4}$

*$\frac{(x^2)(-2y-2x\frac{4-2xy}{x^2}) - ((-2xy+4)(2x))}{x^4}$

*$\frac{(-2x^2y+4x^2y-8x) - (-4x^2y+8x)}{x^4}$

*$\frac{6x^2y-16x}{x^4}$

*$\frac{6(5+4x)-16x}{x^4}$

*$\frac{30+24x-16x}{x^4}$

*$\frac{30+8x}{x^4}$ = $\frac{dy^2}{d^2x}$


Can anyone tell me if I did it correctly?

i have also
$y'=\frac{4-2xy}{x^2}$
and by the quotient rule
$y''=\frac{(-2y-2xy')x^2-(4-2xy)2x}{x^4}$
plugging $y''$ in the equation above we obtain
$y''=\frac{\left(-2y-2x\left(\frac{4-2xy}{x^2}\right)\right)x^2-(4-2xy)2x}{x^4}$
simplifying this we get
$y''=\frac{6xy-16}{x^3}$
