Question

How do you find the equation of tangent line to the curve `y = x^2e^-x`, with the point (1, 1/e)?

Answer 1

`y=x/e`

Explanation:

`y=x^2e^(-x)` can be written in the form of `y=uv` where `u` and `v` are both functions of `x`.

`(dy)/(dx)=vu'+uv'`

`v'=-e^(-x)`

`u'=2x`

`(dy)/(dx)=2xe^(-x)-x^2e^(-x)`

Putting in our `x` value of 1 gives us:

`(dy)/(dx)=2e^(-1)-e^(-1)=e^(-1)`

The gradient is `1/e`

To find the `y`-intercept we put our gradient, an `x`-value and corresponding `y`-value into `y=mx+c`

`1/e=1*1/e+c`

`c=1/e-1/e=0`

`y=x/e`