# How do you find the equation of the tangent line to the graph y=xe^x-e^x through point (1,0)?

Mar 3, 2017

$y = x e - e$

#### Explanation:

Recall that the product rule states that for the function $f \left(x\right) = g \left(x\right) \cdot h \left(x\right)$

$f ' \left(x\right) = g ' \left(x\right) h \left(x\right) + g \left(x\right) h ' \left(x\right)$

For this function, if $y = f \left(x\right) = x {e}^{x} - {e}^{x}$

Factor out ${e}^{x}$
$f \left(x\right) = {e}^{x} \left(x - 1\right)$

Apply the product rule to differentiate
$f ' \left(x\right) = {e}^{x} \left(x - 1\right) + {e}^{x} \left(1\right)$

Simplify
$f ' \left(x\right) = {e}^{x} \left(x - 1 + 1\right) = x {e}^{x}$

Recall that the derivative of a function at a point is equal to the rate of change of that function at that point. In other words, the derivative of $f \left(x\right)$ at $\left(1 , 0\right)$ is equal to the slope of the tangent line at $\left(1 , 0\right)$

Since we have found the derivative function of $f \left(x\right)$, plug in 1 to $f ' \left(x\right)$ to get the slope of the tangent line.

$f ' \left(1\right) = \left(1\right) {e}^{1} = e$

Now, we have the slope of a line and one point that it goes through. We can apply the point-slope form of a line to find its equation.

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

where:
${y}_{1}$ is $f \left(1\right)$, $m = f ' \left(1\right)$, and ${x}_{1} = 1$

Subbing in, we get:

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

$y - 0 = e \left(x - 1\right)$

$y = x e - e$