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.!

How do you use implicit differentiation to find (dy)/(dx)(dy)/(dx) given 2x^3=(3xy+1)^22x^3=(3xy+1)^2?