If #y^3+y=x^2#, what is #(dy)/(dx)#?

Question

10535 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T18:13:21

#dy/dx=(2x)/(3y^2+1)#

#y^3+y=x^2#

By differentiating w.r.t. #x#,

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

By factoring out #dy/dx#,

#Rightarrow (3y^2+1) dy/dx=2x#

By dividing both sides by #(3y^2+1)#,

#Rightarrow dy/dx=(2x)/(3y^2+1)#

I hope that this was clear.