What is the slope of the tangent line of #e^x/(x+y)^3= C #, where C is an arbitrary constant, at #(5,1)#?

1 Answer
Apr 10, 2018

The slope is #1#.

Explanation:

As we are seeking a tangent at #(5,1)#, the curve #e^x/(x+y)^3=C# passes through #(5,1)# and hence #C=e^5/(5+1)^3=e^5/216#

and our curve is #e^x/(x+y)^3=e^5/216#

or #216e^(x-5)=(x+y)^3#

As slope of tangent is given by the value of #(dy)/(dx)# using implicit differentiation of our function #216e^(x-5)=(x+y)^3#

i.e. #216e^(x-5)=3(x+y)^2*(1+(dy)/(dx))#

i.e. #1+(dy)/(dx)=72e^(x-5)/(x+y)^2#

i.e. #(dy)/(dx)=72e^(x-5)/(x+y)^2-1#

and at #(5,1)#, we have

#(dy)/(dx)=72e^(5-5)/6^2-1=2-1=1#

graph{(216e^(x-5)-(x+y)^3)(y-x+4)=0 [-6.04, 13.96, -3.08, 6.92]}