Bear with me here as this will take a while. Whoever gave you this problem clearly does not like you (or challenges you a little too much).
We're finding slope, which means we have to take the derivative of this monstrosity. Taking the derivative will require using the quotient rule, the sum rule, the product rule, and the chain rule. So let's get to it. Taking the derivative of both sides,
#d/dx((x^2y-y^2)/((1+x)^2))=d/dx(C)#
The derivative of a constant, in this case #C#, is 0, which reduces the problem to:
#d/dx((x^2y-y^2)/((1+x)^2))=0#
Now we're left with finding the derivative of this thing. We apply the quotient rule first, which says that:
#d/dx(u/v)=(u'v-uv')/v^2-># where #u# and #v# are functions of #x#...but since we have #y#, we'll use the chain rule to adjust
In our case, #u=x^2y-y^2# and #v=(1+x)^2# (and #v^2=(1+x)^4#). The derivative of #v# is quite simple - using the power rule, we see: #v'=2(1+x)#. The derivative of #u# is a little more complicated. Using the sum rule, we can break #u'# up into: #(x^2y)'-(y^2)'#. The derivative of #x^2y# is found using the product rule, which says:
#d/dx(uv)=u'v+uv'#
Now we define #u=x^2# and #v=y#; it follows that #u'=2x# and #v'=dy/dx#, using the chain rule. The whole derivative is:
#(x^2)'(y)+(x^2)(y)'=2xy+x^2dy/dx#
We have to move on to the other part of the numerator, #y^2#. The derivative with respect to #x# of #y^2# is #2ydy/dx# (using the chain rule); because we don't know what #y# is, we don't know what #dy/dx# is. Multiplying by #dy/dx# is a convenient way of making up for that.
Alright, putting this all together, #u'=(2xy+x^2dy/dx)-(2ydy/dx)#. Using #u#, #u'#, #v#, and #v'#, we are now ready to substitute into the quotient rule formula:
#d/dx((x^2y-y^2)/((1+x)^2))=((x^2y-y^2)'(x+1)^2-(x^2y-y^2)((x+1)^2)')/((x+1)^2)^2=((2xy+x^2dy/dx-2ydy/dx)(x+1)^2-2(x^2y-y^2)(x+1))/((x+1)^4)=((2xy+x^2dy/dx-2ydy/dx)(x+1)-2(x^2y-y^2))/((x+1)^3)#
(The last part there was factoring and canceling a #(x+1)# term)
Time for some algebra:
#0=((2xy+x^2dy/dx-2ydy/dx)(x+1)-2(x^2y-y^2))/((x+1)^3)#
#0=(2xy+x^2dy/dx-2ydy/dx)(x+1)-2(x^2y-y^2)#
#2(x^2y-y^2)=(2xy+x^2dy/dx-2ydy/dx)(x+1)#
#(2(x^2y-y^2))/(x+1)=2xy+dy/dx(x^2-2y)#
#(2(x^2y-y^2))/(x+1)-2xy=dy/dx(x^2-2y)#
#(2(x^2y-y^2)-2xy(x+1))/(x+1)=dy/dx(x^2-2y)#
#(2(x^2y-y^2)-2xy(x+1))/((x+1)(x^2-2y))=dy/dx#
Therefore the slope of the tangent line is given by the expression #(2(x^2y-y^2)-2xy(x+1))/((x+1)(x^2-2y))#. Now all that's left is evaluating this at #(6,1)# to obtain a numerical value:
#dy/dx=(2(x^2y-y^2)-2xy(x+1))/((x+1)(x^2-2y))#
#dy/dx=(2((6)^2(1)-(1)^2)-2(6)(1)((6)+1))/(((6)+1)((6)^2-2(1)))#
#dy/dx=(70-84)/238=-14/238=-7/119=-1/17#
And phew, we're done!
