Derivatives of real functions

Question

• Why don’t endpoints matter?

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Overview §

Definition: Differentiable function on \mathbb R

A function $f:\mathbb{R}\to \mathbb{R}$ is called differentiable at $x_0\in \mathbb{R}$ if:

• (1) There exists a limit

$$f^{\prime }(x_0)=\underset{h\to 0}{\lim }\frac{f(x_0+h)-f(x_0)}{h}.$$

• (2) There exists a linear map $d{f}_{x_0}:\mathbb{R}\to \mathbb{R}$ such that

$$\underset{h\to 0}{\lim }\frac{|f(x_0+h)-f(x_0)-d{f}_{x_0}(h)|}{|h|}=0.$$

In this case, the limit in (1) is called the derivative of $f$ at $x_0$, and denoted $f’(x_0)$. A function is differentiable if it is differentiable at every point.

^DEF-differentiable-function-in-R

Relevant theorems:

• (Theorem) Rolle’s and mean value
• (Theorem) L’Hopital’s rule

Related notes: Local extrema of real functions

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Rules of differential calculus §

Theorem: Sums and products of differentiable functions are differentiable

Let $f,g:\mathbb{R}\to \mathbb{R}$ be differentiable at $x_0\in \mathbb{R}$. Then:

• (1) $(f+g)(x)=f’(x_0)+g’(x_0)$;
• (2) $(fg)(x_0)=f^{\prime }(x_0)g(x_0)+g^{\prime }(x_0)f(x_0)$.

Note that differentiability of $f+g$ and $fg$ follows from the direct computation of the limit.

^THM-sum-product-rule-in-R

Theorem: Power rule for derivatives in \mathbb R

Let $f:\mathbb{R}\to \mathbb{R}$ be defined by $f(x)=x^n$ for some $n\in \mathbb{N}$. Then $f$ is differentiable everywhere and $f’(x)=n{x}^{n-1}$.

^THM-power-rule-in-R

Theorem: Chain rule for derivatives in \mathbb R

Let $f,g:\mathbb{R}\to \mathbb{R}$ be differentiable. Let $x_0\in \mathbb{R}$. Then $g\circ f$ is also differentiable, and

$$(g\circ f)’(x_0)=g’(f(x_0))f’(x_0).$$

^THM-chain-rule-in-R

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Monotone functions and derivatives §

Theorem:

Let $f:[a,b]\to \mathbb{R}$ be differentiable.

• If $f’(x)\ge 0$ for all $x\in [a,b]$, then $f$ is monotonically increasing.
• If $f’(x)=0$ for all $x\in [a,b]$, then $f$ is constant.

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Proof appendix §

Transclude of Derivatives-of-real-functions#^thm-sum-product-rule-in-r

Proof of (2) from Modern Analysis I.

$$\begin{array}{rl}\underset{h\to 0}{\lim }\frac{f(x_0+h)g(x_0+h)-f(x_0)g(x_0)}{h}& =\underset{h\to 0}{\lim }\frac{f(x_0+h)(g(x_0+h)-g(x_0)}{h}\\ & +\underset{h\to 0}{\lim }\frac{g(x_0+h)(f(x_0+h)-f(x_0)}{h}\\ & =f^{\prime }(x_0)g(x_0)+g^{\prime }(x_0)f(x_0)\end{array}$$

Transclude of Derivatives-of-real-functions#^thm-power-rule-in-r

Proof from Modern Analysis I.

$$\begin{array}{rl}\underset{h\to 0}{\lim }\frac{(x_0+h{)}^n-x_0^n}{h}& =\underset{h\to 0}{\lim }\frac{x_0^n+nh{x}_0^{n-1}+\cdots +h^2-x_0^n}{h}\\ & =\lim \frac{nh{x}_0^{n-1}}{h}=\underset{h\to 0}{\lim }n{x}_0^{n-1}=n{x}_0^{n-1}.\end{array}$$

The $\cdots$ are lower-order terms involving $h^2,h^3,\dots ,h^n$.#concept-question try doing this computation explicitly?

Transclude of Derivatives-of-real-functions#^thm-chain-rule-in-r

Proof from Modern Analysis I.

Since $f$ is differentiable, we have

$$f(x+h)=f(x)+f^{\prime }(x)h+\phi (x+h),$$

where $\phi (x)\to 0$ as $x\to x_0$. Similarly,

$$g(y+h)=g(y)+h{g}^{\prime }(y)+\psi (y+h),$$

where $\psi (y+h)\to 0$ as $h\to 0$. We now compute

$$\begin{array}{rl}\underset{h\to 0}{\lim }\frac{g(f(x_0+h))-g(f(x_0))}{h}& =\underset{h\to 0}{\lim }\frac{g(f(x_0)+f^{\prime }(x_0)h+\phi (x_0+h))-g(f(x_0))}{h}\\ & =\underset{h\to 0}{\lim }\frac{g(f(x_0))+h{g}^{\prime }(f^{\prime }(x_0)+\phi (x_0+h))+\psi (f(x_0)+h))-g(f(x_0))}{h}\\ & =g^{\prime }(f(x_0))f^{\prime }(x_0).\end{array}$$