How do you find #(d^2y)/(dx^2)# for #5x^3=-4y^2+4#?
1 Answer
Aug 20, 2016
Explanation:
First, we will use implicit differentiation to find
Differentiating:
#15x^2=-8ydy/dx#
Solve for
#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
#(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)#