How do you implicitly differentiate # xy+2x+3x^2=-4#?

1 Answer
May 13, 2018

So, recall that for implicit differentiation, each term has to be differentiated with respect to a single variable, and that to differentiate some #f(y)# with respect to #x#, we utilise the chain rule:

#d/dx(f(y)) = f'(y)*dy/dx#

Thus, we state the equality:
#d/dx(xy) + d/dx(2x) + d/dx(3x^2) = d/dx(-4)#
#rArr x*dy/dx + y + 2 +6x = 0# (using the product rule to differentiate #xy#).

Now we just need to sort out this mess to get an equation #dy/dx =...#

#x*dy/dx = -6x-2-y#
#:. dy/dx = -(6x+2+y)/x# for all #x in RR# except zero.