Analysis II HT23, Differentitation
Flashcards
@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
\[l = \lim _ {x\to x _ 0} \frac{f(x) - f(x _ 0)}{x - x _ 0} \]exists.
@Define what it means for a function $f$ to be differentiable.
It is differentiable at every point in its domain.
@State what it means for a function $f$ to be differentiable with derivative $l$, _ 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
\[f(x _ 0 + h) = f(x _ 0) + lh + \varepsilon(h)h\]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
\[l = \lim _ {x\to x _ 0} \frac{f(x) - f(x _ 0)}{x - x _ 0} \]
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 $x _ 0$.
@State formally 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 $g \circ f$ is differentiable at $x _ 0$ with
\[(g \circ f)'(x _ 0) = g'(f(x _ 0)) f'(x _ 0)\]@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 differentiation 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
\[l = \lim _ {x\to x _ 0} \frac{f(x) - f(x _ 0)}{x - x _ 0} \]
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
\[f(x _ 0 + h) = f(x _ 0) + lh + \varepsilon(h)h\]
with $\varepsilon(h) \to 0$ as $h \to 0$.
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
\[af'(x _ 0) + bg'(x _ 0)\]
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.
\[\frac{\text d}{\text dx} \frac{f(x)}{g(x)} = \frac{f'(x)g(x)-f(x)g'(x)}{g(x)^2}\]
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 $g \circ f$ is differentiable 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
\[(f^{-1})'(f(x _ 0)) = \frac{1}{f'(x _ 0)}\]
Todo.