Question

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

Answer 1

`y=xe-e`

Explanation:

Recall that the product rule states that for the function `f(x)=g(x)*h(x)`

`f'(x)=g'(x)h(x)+g(x)h'(x)`

For this function, if `y=f(x)=xe^x-e^x`

Factor out `e^x`
`f(x)=e^x(x-1)`

Apply the product rule to differentiate
`f'(x)=e^x(x-1)+e^x(1)`

Simplify
`f'(x)=e^x(x-1+1)=xe^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(x)` at `(1,0)` is equal to the slope of the tangent line at `(1,0)`

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

`f'(1)=(1)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(x-x_1)`

where:
`y_1` is `f(1)`, `m=f'(1)`, and `x_1=1`

Subbing in, we get:

`y-y_1=m(x-x_1)`

`y-0=e(x-1)`

`y=xe-e`