How do you implicitly differentiate #-3=(x^2+y)^3-y^2x #?

1 Answer
Apr 28, 2017

#color(red){dy/dx = [6x(x^2+y)^2 - y^2]/[2xy - 3(x^2+y)^2]}#

Explanation:

Implicit differentiation is basically done in cases where #y# cannot be explicitly written as a function of #x#.

In this case,

#-3=(x^2+y)^3-y^2x #

Differentiating both sides w.r.t. #x#

#=> -3(d(1))/dx = [d(x^2+y)^3]/dx - (d(y^2x))/dx#

Using chain rule to evaluate #[d(x^2+y)^3]/dx# & product rule to evaluate # (d(y^2x))/dx#

#=> 0 = [d(x^2+y)^3]/(d(x^2+y))*[d(x^2+y)]/dx - [y^2dx/dx + xdy^2/dx]#

#=> y^2dx/dx + xdy^2/dx = [d(x^2+y)^3]/(d(x^2+y))*[d(x^2+y)]/dx#

Using sum rule to evaluate #[d(x^2+y)]/dx# & chain rule to evaluate #dy^2/dx#

#=> y^2 + x*dy^2/dy*dy/dx = [3*(x^2+y)^2]*[dx^2/dx + dy/dx]#

#=> y^2 + x*2y*dy/dx = 3(x^2+y)^2*[2x+dy/dx]#

#=> y^2 + 2xy*dy/dx = 6x(x^2+y)^2 + 3(x^2+y)^2*dy/dx#

#=> 2xy*dy/dx - 3(x^2+y)^2*dy/dx = 6x(x^2+y)^2 - y^2 #

#=> [2xy - 3(x^2+y)^2]*dy/dx = 6x(x^2+y)^2 - y^2#

#=> color(red){dy/dx = [6x(x^2+y)^2 - y^2]/[2xy - 3(x^2+y)^2]}#