How do you find the second derivative by implicit differentiation on #x^3y^3=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]#

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

#3x^2y^3 + 3y^2x^3dy/dx = 0#

Now, we will solve for #dy/dx#:

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

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

#dy/dx = -y/x#

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

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

Looking back at the previous equation for #dy/dx#, we can substitute into our equation for the second derivative to get it in terms of only #x# and #y#:

#(d^2y)/(dx^2) = -(x*(-y/x) - y*1)/x^2#

Simplifying yields:

#(d^2y)/(dx^2) = (2y)/x^2#