Question #96976

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Question #96976

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Answer 1 · 2016-03-08T13:06:01

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

Explanation:

$\frac{d}{d x} [x e^{x}] \neq 1 \cdot e^{x}$ $d/dx [xe^x] != 1*e^x$ in a way similar to the way that

$\frac{d}{d x} [x sin x] \neq 1 \cdot sin x$ $d/dx [xsinx] != 1*sinx$ and $\frac{d}{d x} [x \cdot x] \neq 1 \cdot 1$ $d/dx [x*x] != 1 * 1$ and $\frac{d}{d x} [5 x] \neq 0 \cdot 1$ $d/dx [5x] != 0*1$

For differentiable functions $f (x)$ $f(x)$ and $g (x)$ $g(x)$ , 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:

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

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Question #96976

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