# 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 $\left(5 , 1\right)$, the curve ${e}^{x} / {\left(x + y\right)}^{3} = C$ passes through $\left(5 , 1\right)$ and hence $C = {e}^{5} / {\left(5 + 1\right)}^{3} = {e}^{5} / 216$

and our curve is ${e}^{x} / {\left(x + y\right)}^{3} = {e}^{5} / 216$

or $216 {e}^{x - 5} = {\left(x + y\right)}^{3}$

As slope of tangent is given by the value of $\frac{\mathrm{dy}}{\mathrm{dx}}$ using implicit differentiation of our function $216 {e}^{x - 5} = {\left(x + y\right)}^{3}$

i.e. $216 {e}^{x - 5} = 3 {\left(x + y\right)}^{2} \cdot \left(1 + \frac{\mathrm{dy}}{\mathrm{dx}}\right)$

i.e. $1 + \frac{\mathrm{dy}}{\mathrm{dx}} = 72 {e}^{x - 5} / {\left(x + y\right)}^{2}$

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = 72 {e}^{x - 5} / {\left(x + y\right)}^{2} - 1$

and at $\left(5 , 1\right)$, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} = 72 {e}^{5 - 5} / {6}^{2} - 1 = 2 - 1 = 1$

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