How do you find the second derivative by implicit differentiation on x^3y^3=8x3y3=8 ?

1 Answer
Jul 31, 2014

As the first step, we will differentiate once, and apply the product rule:

d/dx[x^3]*y^3 + d/dx[y^3]*x^3 = d/dx[8]ddx[x3]y3+ddx[y3]x3=ddx[8]

For y^3y3, remember to use the chain rule. Simplifying yields:

3x^2y^3 + 3y^2x^3dy/dx = 03x2y3+3y2x3dydx=0

Now, we will solve for dy/dxdydx:

dy/dx = -(3x^2y^3)/(3y^2x^3)dydx=3x2y33y2x3

We can cancel off the 3, an x^2x2, and a y^2y2, which will yield:

dy/dx = -y/xdydx=yx

Now, differentiate once again. We will apply the quotient rule:

(d^2y)/(dx^2) = -(x*dy/dx - y*1)/x^2d2ydx2=xdydxy1x2

Looking back at the previous equation for dy/dxdydx, we can substitute into our equation for the second derivative to get it in terms of only xx and yy:

(d^2y)/(dx^2) = -(x*(-y/x) - y*1)/x^2d2ydx2=x(yx)y1x2

Simplifying yields:

(d^2y)/(dx^2) = (2y)/x^2d2ydx2=2yx2