How do you use implicit differentiation to find #(dy)/(dx)# given #2x^3=(3xy+1)^2#?

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Answer 1 · 2016-09-27T13:29:20

#dy/dx={x^2-y(3xy+1)}/{x(3xy+1)}#.

#2x^3=(3xy+1)^2#.

Diff.ing both sides w.r.t. x,

#d/dx(2x^3)=d/dx(3xy+1)^2#.

Here, by the Chain Rule,

#"The R.H.S.="d/dx(3xy+1)^2=2(3xy+1)d/dx(3xy+1)#

#=2(3xy+1){d/dx(3xy)+d/dx(1)}#

#=2(3xy+1){3d/dx(xy)+0}#

#=2(3xy+1)[3{xd/dx(y)+yd/dx(x)}]........"[Product Rule]"#

#=2(3xy+1){3(xdy/dx+y)}#

#=6(3xy+1)(xdy/dx+y).....................(1)#

#"The L.H.S.="d/dx(2x^3)=2*3x^2=6x^2...................(2)#

From #(1) and (2)#, we have,

#cancel(6)x^2=cancel(6)(3xy+1)(xdy/dx+y)", i.e.,"#

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

#:. {x^2-y(3xy+1)}/{x(3xy+1)}=dy/dx#.

Enjoy Maths.!