What does it mean for a function $f(x)$ to be differentiable at $x_0$?
I need this to understand more concepts in real-analysis and calculus.
Thank you.
$\endgroup$ 34 Answers
$\begingroup$By definition, if x is a point in the domain of a function f, then f is said to be differentiable at x if the derivative f′(x) exists. Intuitively, this means that the graph of f has a non-vertical tangent line at the point (x, f(x)). This also means that the function, at the point x, has no bends, cusps, breaks, or, as mentioned, a vertical tangent.
$\endgroup$ 1 $\begingroup$That a function $f(x)$ is differentiable at $x_0$ means that the derivative of $f(x)$ evaluated at $x_0$ exists, which means that the following exists: $$\lim\limits_{h\to0}\frac{f(x_0+h)-f(x_0)}{h}=f'(x_0).$$Examples:
Let $f(x)=x^{-1}$, then $f'(x)=-x^{-2}$. We say $f(x)$ is differentiable at $3$ because $f'(3)=-1/9$ exists. We say $f(x)$ is not differentiable at $0$ since division by zero is undefined and thus $f'(0)$ doesn't exist.
I hope this helps.
Best wishes, $\mathcal H$akim.
Intuitively, a function is differentiable if it is locally linearly approximable. That means for a given point $x_0$ in the domain of $f$, there is a linear approximation of $f$ near $x_0$. The error of the linear approximation falls to zero as you restrict the approximation to smaller and smaller neighborhoods of the point $x_0$ (that's what "approximable" means).
For functions $\mathbb{R}\rightarrow\mathbb{R}$, the graph of the approximating function is a small piece of the tangent to the graph of $f$ at $x_0$.
$\endgroup$ $\begingroup$for example
$f(x)=x^2$ is differentiable at $x=0$
but $f(x)=1/x$ is not differentiable at $x=0$ because $f'(x)=-1/x^2$
which is not defined at $x=0$
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