# Notes - Analysis II HT23, Differentitation

### Flashcards

Can you state what it means for a function $f$ to be differentiable with derivative $l$ (with a division involved)?

Let $f : E \to \mathbb R$ be a function and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ if

\[l = \lim_{x\to x_0} \frac{f(x_0) - f(x)}{x - x_0}\]exists.

What does it mean for a function $f$ to be differentiable?

It is differentiable at every point in its domain.

Can you state what it means for a function $f$ to be differentiable with derivative $l$, *without using division in the definition*?

*without using division in the definition*?

Let $f : E \to \mathbb R$ and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ and $f’(x _ 0) = l$ if and only if one can write

\[f(x_0 + h) = f(x) + lh + \varepsilon(h)h\]with $\varepsilon(h) \to 0$ as $h \to 0$.

Quickly justify that the two following characterisations of $f$ being differentiable at $x _ 0$ are equivalent:

Let $f : E \to \mathbb R$ be a function and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ if

\[l = \lim_{x\to x_0} \frac{f(x_0) - f(x)}{x - x_0}\]
and

Let $f : E \to \mathbb R$ and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ and $f’(x _ 0) = l$ if and only if one can write

\[f(x_0 + h) = f(x_0) + lh + \varepsilon(h)h\]
with $\varepsilon(h) \to 0$ as $h \to 0$.

Note that if $h \ne 0$, then

\[\varepsilon (h) = \frac{f(x_0+h) -f(x_0)}{h} - l\]and so the following are equal if and only if $\varepsilon (h) \to 0$ as $h \to 0$.

What implication links continuity and differentiability for $f$ at $x _ 0$?

If $f$ is differentiable at $x _ 0$, then $f$ is continuous at $x _ 0$.

Can you formally state what is meant by the chain rule for functions $f$ and $g$?

Suppose $f : E \to \mathbb R$ and, $g : E’ \to \mathbb R$ are functions with $f(E) \subseteq E’$. Then if

- $f$ is differentiable at $x _ 0 \in E$,
- $g$ is differentiable at $f(x _ 0) \in E’$

then $g \circ f$ at $x _ 0$ with

\[(g \circ f)'(x_0) = g'(f(x_0)) f'(x_0)\]Can you state the inverse function theorem (for derivatives)?

Let $f : I \to \mathbb R$ where $I$ is an interval, $f$ is strictly monotonic, and that $f$ is differentiable at $x _ 0$ with non-zero derivative. Then $f^{-1} : f(I) \to I$ is differentiable at $f(x _ 0)$ with

\[(f^{-1})'(f(x_0)) = \frac{1}{f'(x_0)}\]When proving the linearity of differentitation and the product rule for $f(x), g(x)$ what characterisation of differentiation do you use?

When proving the quotient rule for differentiation, you start by considering just $\frac{1}{g’(x)}$. What characterisation of differentiation do you use here and how do you get started?

### Proofs

Prove that the following two definitions of what it means for $f$ to be differentiable at $x _ 0$ are equivalent:

(Division-y) Let $f : E \to \mathbb R$ be a function and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ with derivative $l$ if

\[l = \lim _ {x\to x _ 0} \frac{f(x _ 0) - f(x)}{x - x _ 0}\]
exists.

and

(Division-less) Let $f : E \to \mathbb R$ and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ and $f’(x _ 0) = l$ if one can write

\[f(x _ 0 + h) = f(x) + lh + \varepsilon(h)h\]
with $\varepsilon(h) \to 0$ as $h \to 0$.

(Division-y) Let $f : E \to \mathbb R$ be a function and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ with derivative $l$ if

(Division-less) Let $f : E \to \mathbb R$ and let $x _ 0 \in E$ be a limit point of $E$. Then $f$ is differentiable at $x _ 0$ and $f’(x _ 0) = l$ if one can write

Todo.

Prove that if $f$ is differentiable at $x _ 0$, then $f$ is continuous at $x _ 0$.

Todo.

Prove the linearity of the derivative, i.e. if

\[af(x) + bg(x)\]
is differentiable at $x _ 0$, then it has derivative

\[af'(x_0) + bf'(x_0)\]

Todo.

Prove the product rule, i.e.

\[\frac{\text d}{\text dx} f(x)g(x) = f'(x) g(x) + f(x) g'(x)\]

Todo.

Prove the quotient rule from the definition of derivatives, i.e.

\[\frac{\text d}{\text dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x)-f(x)g'(x_0)}{g(x_0)^2}\]

Todo.

Prove the chain rule for differentiation:

Suppose $f : E \to \mathbb R$ and, $g : E’ \to \mathbb R$ are functions with $f(E) \subseteq E’$. Then if

- $f$ is differentiable at $x _ 0 \in E$,
- $g$ is differentiable at $f(x _ 0) \in E’$

then $g \circ f$ at $x _ 0$ with

\[(g \circ f)'(x_0) = g'(f(x_0)) f'(x_0)\]

Todo.

Prove the inverse function theorem for derivatives:

Let $f : I \to \mathbb R$ where $I$ is an interval, $f$ is strictly monotonic, and that $f$ is differentiable at $x _ 0$ with non-zero derivative. Then $f^{-1} : f(I) \to I$ is differentiable at $f(x _ 0)$ with

\[(f^{-1})'(f(x _ 0)) = \frac{1}{f'(x _ 0)}\]

Let $f : I \to \mathbb R$ where $I$ is an interval, $f$ is strictly monotonic, and that $f$ is differentiable at $x _ 0$ with non-zero derivative. Then $f^{-1} : f(I) \to I$ is differentiable at $f(x _ 0)$ with

Todo.