How do you find #(d^2y)/(dx^2)# for #5x^3=-4y^2+4#?

1 Answer
Aug 20, 2016

#(d^2y)/dx^2=-(240xy^2+225x^4)/(64y^3)#

Explanation:

First, we will use implicit differentiation to find #dy/dx#. Recall that implicit differentiation involves using the chain rule when differentiating a term with #y#.

Differentiating:

#15x^2=-8ydy/dx#

Solve for #dy/dx#:

#dy/dx=-(15x^2)/(8y)#

Differentiate again. This time the quotient rule will be used!

#(d^2y)/dx^2=-(30x(8y)-15x^2(8dy/dx))/(8y)^2#

Replace #dy/dx# with #-(15x^2)/(8y)#:

#(d^2y)/dx^2=-(30x(8y)-15x^2(8(-(15x^2)/(8y))))/(8y)^2#

Simplify:

#(d^2y)/dx^2=-(240xy-15x^2(-(15x^2)/y))/(64y^2)#

#(d^2y)/dx^2=-(240xy^2-15x^2(-15x^2))/(64y^3)#

#(d^2y)/dx^2=-(240xy^2+225x^4)/(64y^3)#