Question

Question #96976

Answer 1

`xe^x` is differentiated using the product rule because it is a product of two functions. (The fact that the derivative of `e^x` is also `e^x` does not change the fact that `xe^x` is a product.)

Explanation:

`d/dx [xe^x] != 1*e^x` in a way similar to the way that

`d/dx [xsinx] != 1*sinx` and `d/dx [x*x] != 1 * 1` and `d/dx [5x] != 0*1`

For differentiable functions `f(x)` and `g(x)`, the derivative of the product is

`d/dx[f(x)g(x)] = lim_(hrarr0) ([f(x+h)g(x+h)]-[f(x)g(x)])/h`

`= lim_(hrarr0) ([f(x+h)g(x+h)]- [f(x)g(x+h)] + [f(x)g(x+h)]- [f(x)g(x)])/h`

`= lim_(hrarr0) ([f(x+h)g(x+h)]- [f(x)g(x+h)])/h + ([f(x)g(x+h)]- [f(x)g(x)])/h`

`= lim_(hrarr0) (f(x+h)- f(x)]/h g(x+h) + f(x)(g(x+h) - g(x))/h`

`= lim_(hrarr0) (f(x+h)- f(x)]/h lim_(hrarr0)g(x+h) + lim_(hrarr0)f(x) lim_(hrarr0)(g(x+h) - g(x))/h`

`= d/dx[f(x)] g(x)+f(x)d/dx[g(x)]`

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

Applied to `xe^x`

`d/dx[xe^x] = d/dx(x) e^x + x d/dx(e^x)`

`= 1*e^x + x e^x`

In the function for this question, we have a quotient `f(x) = u/v` with

`u = 1-xe^x` and `v = x+e^x`.

We'll be using the quotient rule, so we need

`u' = 0-[1e^x+xe^x] = -(e^x+xe^x)` and `v' = 1+e^x`

`f'(x) = (u'v-uv')/v^2 = (-(e^x+xe^x)(x+e^x) - (1-xe^x)(1+e^x))/(x+e^x)^2`

`= (-[(e^x+xe^x)(x+e^x) + (1-xe^x)(1+e^x)])/(x+e^x)^2`

Expand and simplify to get:

`f'(x) = (-[x^2e^x+e^x+e^(2x)+1])/(x+e^x)^2`