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

Question

1823 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-29T05:38:09

#(d^2y)/(dx^2)=(15x)/(4y)-(225x^4)/(64y^3)#

As #5x^3=4y^2+4#, differentiating

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

i.e. #(dy)/(dx)=(15x^2)/(8y)#

differentiating (A) further, we get

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

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

or #8y(d^2y)/(dx^2)=30x-(225x^4)/(8y^2)#

or #(d^2y)/(dx^2)=(30x)/(8y)-(225x^4)/(64y^3)#

#=(15x)/(4y)-(225x^4)/(64y^3)#