Question

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

Answer 1

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]}