Differentiability

A function is said to be differentiable if a derivative exists for it over a given domain or a point.#todo Why is it important to know if a function is differentiable?

Definition §

If $f$ is a real-valued function, then the derivate of $f$ at $x=a$ is defined by:

$f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}$ which is derived from Derivation by First Principles.

$f$ is said to be differentiable at x = a iff this limit exists.

Theorem - Differentiability and Continuity §

If $f$ is differentiable at $x=a$, then $f$ is also continuous at $x=a$:

$f^{\prime }(a)={\lim }_{h\to 0}\frac{f(a+h)-f(a)}{h}⟹{\lim }_{x\to a}f(x)=f(a)$ Using the contrapositive, discontinuity implies that a function is non-differentiable. I.e. if $f$ is not continuous over a point, then it is not differentiable either.

L’Hôpital’s Rule §

Definition §

Let $f,g$ be real-valued functions.

${\lim }_{x\to a}\frac{f(x)}{g(x)}={\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$

• The function must be in the form $\frac{\infty }{\infty }$ or $\frac{0}{0}$ .
• Given an open interval ((3,4), (a,b), etc.) $D$ where $x\ne a$:
  • $f$ and $g$ must be differentiable in $D$
  • $g^{\prime }(x)\ne 0$ in $D$
• Obviously, ${\lim }_{x\to a}\frac{f^{\prime }(x)}{g^{\prime }(x)}$ needs to exist

The rule also works for one-sided limits.

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