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The function I come up with is:

$f(x)= -5x$

and I think this function can be differentiated everywhere because the domain is $\mathbb R$ right?

$f'(x) = -5$ which is constant, but my question is that how can I use theorem to prove my answer.

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    I think you're confused by something. The function $-5x$ has derivative $-5$ at every point of $\mathbb R.$You're done. – 2017-01-05

1 Answers 1

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To show that $f$ is differentiable at all $x \in \Bbb R$, we must show that $f'(x)$ exists at all $x \in \Bbb R$.

Recall that $f$ is differentiable at $x$ if $\lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h}$ exists.

So for $f(x)=-5x$, we examine $$\lim\limits_{h \to 0} \frac{-5(x+h)- (-5x)}{h} = \lim\limits_{h \to 0} \frac{-5h}{h} = \lim\limits_{h \to 0}{-5} = -5$$ And so we see that $f$ is differentiable at all $x \in \Bbb R$ with derivative $f'(x)= -5$.

We could also say that if $g(x)$ and $h(x)$ are differentiable, then so too is $f(x) = g(x) h(x)$ and that $f'(x) = g'(x) h(x) + g(x) h'(x)$. Then let $g(x) = x, \; h(x) = -5$, noting that both are differentiable with derivates $1$ and $0$ respectively, leading to the same result.