03. Differentiability and Derivatives

$\gdef \N{\mathbb{N}} \gdef \Z{\mathbb{Z}} \gdef \Q{\mathbb{Q}} \gdef \R{\mathbb{R}} \gdef \C{\mathbb{C}} \gdef \setcomp#1{\overline{#1}} \gdef \sseq{\subseteq} \gdef \pset#1{\mathcal{P}(#1)} \gdef \covariant{\operatorname{Cov}} \gdef \of{\circ} \gdef \p{^{\prime}} \gdef \pp{^{\prime\prime}} \gdef \ppp{^{\prime\prime\prime}} \gdef \pn#1{^{\prime\times{#1}}} $

May 8 Lecture. Links to the board and the recording.
What does this stuff do?
If you have a function \(y=f(x)\), we want to be able to calculate the slope of a line that is tangent to a point on the graph.
We seek a derivative function \(f^\prime(x)\) such that
\(f^\prime(x_0)=\) slope of line tangent to graph \((y=f(x))\) at the point \((x_0,f(x_0))\).

Why are we interested, and how do we find this?
Background:
What is a tangent line to a graph?
Let’s take the following example:

Let \(x=time\).
Let \(f(x) =\) distance traveled by time \(x\).
Speed = \(\frac{distance}{time}\).
If \(f(2)=1\), \(f(3)=4\), then my speed over time interval \([2,3]\) is \(\frac{f(3)-f(2)}{3-2}\).
I.E. \(\frac{\Delta y}{\Delta x} = 3 \Rightarrow \text{ slope of line connecting } (2, f(2)) \text{ and } (3, f(3)).\)
Slope represents the rate of change as in this example.

What would the speedometer show at \(x=3\)?
Idea: let’s take shorter and shorter intervals around \(x=3\) and see what the slope for that line is. Each slope is a rate of change (ROC) of \(f\) over a smaller and smaller interval \([x,3]\). Define the Instantaneous Rate of Change (IRoC) at \(x=3\) to be the limit of these slopes.

Or, if we express \(x=x_0 + h\), IRoC at \(x_0 = \lim_{h\to 0}{\frac{f(h_0+h)-f(x_0)}{h}}\)
The tangent line to \(y=f(x)\) at \(x=x_0\) is defined as the line through \((x_0, f(x_0))\) with this slope.

Definitions:¶

1. \(f(x)\) is differentiable at the point \(x=x_0\) if the limit \(\lim_{h\to 0}{\frac{f(x_0+h)-f(x_0)}{h}}\) exists.
2. \(f(x)\) is differentiable from the left if \(\lim_{h\to 0^-}{\frac{f(x_0+h)-f(x_0)}{h}}\) exists.

3. \(f(x)\) is differentiable from the right if \(\lim_{h\to 0^+}{\frac{f(h_0+h)-f(x_0)}{h}}\) exists.

   • \(f\) is differentiable at \(x_0\) \(\iff\) \(f\) is differentiable from left and from right at \(x_0\) and these limits are equivalent.

   • Example: Graph \(y=|x|\). Is x differentiable at \(x=0\)?
     From the left: \(\lim_{h\to 0^-}{\frac{f(h)-f(0)}{h}}\) = \(\lim_{h\to 0}{\frac{-h}{h}}=-1\). Note, \(h<0\), so \(|h|=-h\)
     From the right: \(\lim_{h\to 0^+}{\frac{f(h)-f(0)}{h}}=1\)
     Hence, \(f(x)=|x|\) is differentiable from each side at \(x_0=0\), but is not differentiable there.

4. \(f\) is differentiable in open interval \((a,b)\) if f is differentiable at each pt. in that interval

5. If \(f\) is differentiable in \((a,b)\) then the derivative of \(f\) on \((a,b)\) is defined as (for \(x_0\in(a,b)\)): \(f^\prime(x_0)=\lim_{h\to 0}{\frac{f(x_0+h)-f(x_0)}{h}}\)

One sided derivatives are defined similarly to definitions 2 & 3.

The derivative function \(f^\prime\) is defined for any \(x_0\) in this manner:

Theorem 31:¶

Linear Approximation using derivatives:¶

If one would want to calulate \((1.0123)^2\)

\(f^\prime(x)=\lim_{h\to 0}{\frac{f(x+h)-f(x)}{h}}\).

I.e., for small \(h\): \(f^\prime (x) \approx \frac{f(x+h)-f(x)}{h}\)

For smaller h, the approximation is more exact.

\(f(x+h)\approx f(x)+h\cdot f^\prime (x)\).

(We multiply by 2, because the derivative of \(f\) is \(2x\))

Differentiation Rules¶

1. Derivative of a constant function \(f(x)=c\) is zero.
   • \(f(x)=c\) is differentiable everywhere, and \(f^\prime(x)=0\)
2. If \(f\) and \(g\) are differentiable at \(x_0\), so is the function \(f+g\), and \((f+g)\prime(x_0)\) = \(f^\prime(x_0) + g\prime(x_0)\).
   • The derivative of a sum of functions at a point is the sum of the derivatives of functions at a point.
3. if f is differentiable at \(x=x_0\) and c is any real number, then cf is diff. at \(x_0\), and \({(c\cdot f)}^\prime = c\cdot f^\prime\).

May 8 Tirgul. Links to the board and the recording.

Some Derivative Functions¶

\(f(x)=x^3\) (in general, \(f(x) = x^n\) for \(n\in\N\))
\(f^\prime(x)=\lim_{h\to0}{\frac{ {(x+h)}^3-x^3}h}\)
\(=\lim_{h\to0}{\frac{x^3+3x^2h+3xh^2+h^3-x^3}{h}}\)
\(=\lim_{h\to0}{\frac{3x^2h+3xh^2+h^3}{h}}\)
\(=\lim_{h\to0}{\frac{h\cdot(3x^2+3xh+h^2)}{h}}\)
\(=\lim_{h\to0}{3x^2+3xh+h^2}\)
\(=3x^2 \Rightarrow f^\prime(x)=3x^2\)

In general
if \(f(x) = x^n\) for \(n\in\N\),

E.g. \(f(x) = x^5\Rightarrow f^\prime(x)=5x^4\)

May 10 Lecture. Links to the board and the recording.

Rules of Differentiation:¶

1. \(f(x)=c\Rightarrow f^\prime(x)=0\)
2. \({(c\cdot f)}^\prime(x)=c\cdot f^\prime(x)\)
3. \({(f\pm g)}^\prime(x)=f^\prime(x)\pm g^\prime(x)\)
4. \({(f\cdot g)}^\prime(x)=f(x)\cdot g^\prime(x)+f^\prime(x)\cdot g(x)\)
   Product Rule
5. \({(\frac fg)}^\prime(x)=\frac{f^\prime(x)\cdot g(x)-f(x)\cdot g^\prime(x)}{{\left[g(x)\right]}^2}\)
   Quotient Rule

Chain Rule¶

Used to differentiate composed function \(f(g(x))\)
Definition:

Applications of the Chain Rule:¶

1. \({(x^{\frac pq})}^\prime=\frac pq x^{\frac pq -1}\)

2. Given \(f(x)\), (who’s derivative \(f^\prime\) we know), what is derivative of \(f^{-1}\)?

   • We find formula here, this is known as the Inverse Function Rule:
   \[{\left[f^{-1}(x)\right]}^\prime=\frac1{f^\prime(f^{-1}(x))}\]
   • Ex. \(f(x)=\ln x\Rightarrow f^\prime(x)=\frac1x\qquad \text{What's derivative of }f^{-1}(x)=e^x\)?
     • \({(e^x)}^\prime=\frac1{\frac1{e^x}}=e^x\)
   • \(f(x)=\tan x\Rightarrow f^\prime(x)=\frac1{\cos^2x}\qquad \text{What's derivative of }f^{-1}(x)=\arctan x\)?
     • \({(\arctan x)}^\prime=\frac1{\frac1{\cos^2(\arctan x)}}=\cos^2(\arctan x)\Rightarrow\)lots of math\(\Rightarrow\frac1{x^2+1}\)
   • \(f(x)=x^2\Rightarrow f^\prime(x)=2x\qquad \text{What's derivative of }f^{-1}(x)=\sqrt x\)?
     • \({(\sqrt x)}^\prime=\frac1{2\sqrt x}\)

May 15 Lecture. Links to the board and the recording.

Other useful derivatives:¶

\(f(x)=2^x\) (in general: \(f(x)=a^x\mid (a>0)\))
Recall: \(2=e^{\ln2}\Rightarrow2^x={(e^{\ln2})}^x=e^{x\ln2}\)
This is a composed function \((g\of h)(x)\) where \(h(x)=x\cdot\ln2\) and \(g(x)=e^x\).
So, we apply the Chain Rule to get \(e^{x\cdot\ln2}\cdot\ln2\Rightarrow\ln2\cdot 2^x\)

In general:

\(f(x)=a^x\Rightarrow f^\prime(x)=(\ln a)\cdot a^x\)
\(f(x)=\log_2(x)\) (in general: \(f(x)=\log_a(x)\mid(a>0,a\not=1)\))
Use the Inverse Function Rule:
Let \(f(x)=a^x\), so that \(f^{-1}(x)=\log_ax\).
\({(\log_ax)}^\prime=\frac1{f^\prime(f^{-1}(x))}=\frac1{\ln a\cdot a^{\log_ax}}=\frac1{\ln a\cdot x}\)
\(f(x)=|g(x)|\)
\(f^\prime(x)=\frac {g(x)}{|g(x)|}\cdot g^\prime(x)\)
\(f(x)=\ln(g(x))\)
\(f^\prime(x)=\frac{1}{g(x)}\cdot g^\prime(x)=\frac{g^\prime(x)}{g(x)}\)

Trigonometric derivatives:¶

Second Derivative¶

Definition: The second derivative of \(f\) at \(x=a\) is the derivative of \(f^\prime(a)\), evaluated at \(x=a\).
We’ll learn later the graphical significance of \(f^{\prime\prime}(x)\).

In “physical” terms, if \(f(x)\) denotes distance, then \(f^\prime(x)\) denotes velocity (change of distance over time), and \(f^{\prime\prime}(x)\) denotes acceleration (change of velocity over time)

May 15 Tirgul. Links to the board and the recording.
[Incomplete]

May 17 Lecture. Links to the board and the recording.
[Incomplete]