Question

Find (d^2)y/d(x^2) in terms of x and y of (x^2)(y^2) - 6 x = 6

using implicit differentiation


(2(x^2)(y^4)-3x(y^2)-3)/((x^4)(y^3) is not the right answer and i'm not sure y.

Answer 1

implicit differentiation one time

then solve for dy/dx
then take derivitive again
I will explain later


first time
dy/dx
$2xy^2+2yx^2 \space\ \frac{dy}{dx}-6=0$
solve for $\frac{dy}{dx}$
add 6 to both sides and minus 2xy²
$2yx^2 \space\ \frac{dy}{dx}=6-2xy^2$
divide both sides by 2yx²
$\frac{dy}{dx}=\frac{6-2xy^2}{2yx^2}$
$\frac{dy}{dx}=\frac{3-xy^2}{xy^2}$

now we do it again
but this time, if you take the derivitive of y, we replace it with dy/dx or $\frac{3-xy^2}{xy^2}$

use quotient rule

$\frac{dy^2}{dx^2}=\frac{(-2xy^3 \space\ \frac{dy}{dx})(xy^2)-(2xy^3 \space\ \frac{dy}{dx})(3-xy^2)}{x^2y^6}$
$\frac{dy^2}{dx^2}=\frac{(-2xy^3)(\frac{3-xy^2}{xy^2})(xy^2)-(2xy^3)(\frac{3-xy^2}{xy^2})(3-xy^2)}{x^2y^6}$
$\frac{dy^2}{dx^2}=\frac{(-2xy^3)(3-xy^2)-(2xy^3)(\frac{3-xy^2}{xy^2})(3-xy^2)}{x^2y^6}$
$\frac{dy^2}{dx^2}=\frac{(-6xy^2+2x^2y^5)-(\frac{6xy^2-2x^2y^5}{xy^2})(3-xy^2)}{x^2y^6}$
$\frac{dy^2}{dx^2}=\frac{(-6xy^2+2x^2y^5)-(\frac{18xy^2-6x^2y^5-6x^2y^4+2x^3y^7}{xy^2})}{x^2y^6}$
just simplify because I don't have my calculator right now