How do you differentiate #x * y + 2x + 3x^2 = 4#?

1 Answer
Aug 3, 2015

#d/dx[xy+2x+3x^2=4]# results in #dy/dx=3x-4-6/x#

Explanation:

Using Implicit Differentiation we have

#d/dx[xy+2x+3x^2=4]#
#d/dx[xy]+d/dx[2x]+d/dx[3x^2]=0#
#x*d/dx[y]+y*d/dx[x]+2*d/dx[x]+3*d/dx[x^2]=0#
#x*y'+y*1+2*1+3*(2x)=0#
#x*y'+y=-6x-2#
#y'=(-6x-2)/x-y#

Solving for y in the original problem, we have

#x*y+2x+3x^2=4#
#x*y=4-2x-3x^2#
#y=(4-2x-3x^2)/x#

Now, plug this value for #y# into the equation for #y'# above

#y'=(-6x-2)/x-y#
#y'=(-6x-2)/x-(4-2x-3x^2)/x#
#y'=(-6x-2-4+2x+3x^2)/x#
#y'=(-4x-6+3x^2)/x#
#y'=(3x^2-4x-6)/x#
#y'=3x-4-6/x#