What is the slope of the tangent line of # 3y^2+4xy+x^2y =C #, where C is an arbitrary constant, at #(2,5)#?

1 Answer
Jul 18, 2016

#dy/dx=-20/21#

Explanation:

You will need to know the basics of implicit differentiation for this problem.

We know the slope of the tangent line at a point is the derivative; so the first step will be to take the derivative. Let's do it piece by piece, starting with:
#d/dx(3y^2)#

This one isn't too hard; you just have to apply the chain rule and power rule:
#d/dx(3y^2)#
#->2*3*y*dy/dx#
#=6ydy/dx#

Now, onto #4xy#. We will need the power, chain, and product rules for this one:
#d/dx(4xy)#
#->4d/dx(xy)#
#=4((x)'(y)+(x)(y)')-># Product rule: #d/dx(uv)=u'v+uv'#
#=4(y+xdy/dx)#
#=4y+4xdy/dx#

Alright, finally #x^2y# (more product, power, and chain rules):
#d/dx(x^2y)#
#=(x^2)'(y)+(x^2)(y)'#
#=2xy+x^2dy/dx#

Now that we have found all of our derivatives, we can express the problem as:
#d/dx(3y^2+4xy+x^2y)=d/dx(C)#
#->6ydy/dx+4y+4xdy/dx+2xy+x^2dy/dx=0#
(Remember the derivative of a constant is #0#).

Now we collect terms with #dy/dx# on one side and move everything else to the other:
#6ydy/dx+4y+4xdy/dx+2xy+x^2dy/dx=0#
#->6ydy/dx+4xdy/dx+x^2dy/dx=-(4y+2xy)#
#->dy/dx(6y+4x+x^2)=-(4y+2xy)#
#->dy/dx=-(4y+2xy)/(6y+4x+x^2)#

All that's left to do is plug in #(2,5)# to find our answer:
#dy/dx=-(4y+2xy)/(6y+4x+x^2)#
#dy/dx=-(4(5)+2(2)(5))/(6(5)+4(2)+(2)^2)#
#dy/dx=-(20+20)/(30+8+4)#
#dy/dx=-(40)/(42)=-20/21#