Question

What is the equation of the line tangent to `f(x)= xe^x` at `x=7`?

Answer 1

You first must differentiate by the product rule.

Explanation:

Let `f(x) = g(x) xx h(x)`, then `g(x) = x` and `h(x) = e^x`. The product rule states that `f'(x) = g'(x) xx h(x) + g(x) xx h'(x)`

The derivative of `x` is `1`. The derivative of `e^x` is `e^x`.

Hence, `f'(x) = 1 xx e^x + xe^x =e^x + xe^x = e^x(1 + x)`

The function and the tangent pass through the point `(7, 7e^7)`.

All that we have left to find is the slope of the tangent. The slope of the tangent is given by evaluating `f'(a)`, where `x = a` is the given point.

`m_"tangent" = e^7(1 + 7) = 8e^7`

Now, by point-slope form we have:

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

`y - 7e^7 = 8e^7(x - 7)`

`y - 7e^7 = 8e^7x - 56e^7`

`y = 8e^7x - 49e^7`

In summary, the equation of the tangent of `f(x) = xe^x` at the point `x = 7` is `y = 8e^7x - 49e^7`.

Hopefully this helps!