MIT OCW Single Variable Calculus - Week 2

Week 2

Derivative Formulas

To work towards the goal of “differentiating everything”, need to talk about derivative formulas.

• Specfic – \(f’(x)\), where \(f(x) = x^n, \frac{1}{x}\) as a few examples.
• General – \((u + v)’ = u’ + v’\) and \((cu)’ = cu’\) where c is constant, as a few examples.

Start with specific trigonometric derivative formulas.

\(\frac{\delta}{\delta x} sin \; x\)

\[
\frac{\delta}{\delta x} sin \; x = \lim_{x \to 0} \frac{sin(x + \Delta) \; – sin(x)}{\Delta x}
\]

To solve this, we need to revist the trigonometric sum and difference formulas, namely:

\[
sin(A + B) = sin A \; cos B + cos A \; sin B
\]

This is a proof of the identity, using Eulers formula

Going back to the derivative:

\[
\begin{align}
\frac{\delta}{\delta x} sin \; x &= \lim_{x \to 0} \frac{sin(x + \Delta) \; – sin(x)}{\Delta x} \\
&= \lim_{x \to 0} \frac{sin\:x \; cos\:\Delta x + cos\;x \; sin\:\Delta x – sin\:x}{\Delta x} \\
&= \lim_{x \to 0} \quad sin\:x(\frac{cos\:\Delta x – 1}{\Delta x}) + cos\:x(\frac{sin\:\Delta x}{\Delta x}) \\
&= cos\:x
\end{align}
\]

due to the fact that \(\frac{cos\:\Delta x – 1}{\Delta x}\) tends to 0 as \(\Delta x\) tends to 0, and \(\frac{sin\:\Delta x}{\Delta x}\) tends to 1 as \(\Delta x\) tends to 0. Label these as A and B respectively, as we will prove them later.

\(\frac{\delta}{\delta x} cos x\)

Same as above, only this time using the identity:

\[
cos(A + B) = cos A \; cos B – sin A \; sin B
\]

\[
\begin{align}
\frac{\delta}{\delta x} cos \; x &= \lim_{x \to 0} \frac{cos(x + \Delta) \; – cos(x)}{\Delta x} \\
&= \lim_{x \to 0} \frac{cos\;x \: cos\;\Delta x – sin\;x \: sin\;\Delta x – cos\;x}{\Delta x} \\
&= \lim_{x \to 0} \quad cos\;x(\frac{cos\;\Delta x – 1}{\Delta x}) – sin\;x(\frac{\sin\;\Delta x}{\Delta x}) \\
&= \quad – sin\;x
\end{align}
\]

again, by A and B as stated above.

Proofs of A and B

First, some remarks:

\[
\frac{\delta}{\delta x} cos\;x \quad \Bigg|^{x=0}
\quad \lim_{\Delta x \to 0} \frac{cos(\Delta x) – 1}{\Delta x}
\]

This shows that the rate of change of \(cos x\) at \(x = 0\) is a limit as \(\Delta x\) tends to 0, which has been simplified:

\[
\begin{align}
&\frac{cos(0 + \Delta x) – cos(0)}{\Delta x} \\
= &\frac{cos(\Delta x) – 1}{\Delta x}
\end{align}
\]

and likewise

\[
\frac{\delta}{\delta x} sin\;x \quad \Bigg|^{x=0}
\quad \lim_{\Delta x \to 0} \frac{sin(\Delta x)}{\Delta x}
\]

because of the simplification:

\[
\begin{align}
&\frac{sin(0 + \Delta x) – sin(0)}{\Delta x} \\
= &\frac{sin(\Delta x)}{\Delta x}
\end{align}
\]

We should note that the derivatives of sine and cosine at \(x = 0\) give all values of \(\frac{\delta}{\delta x} sin x\), \(\frac{\delta}{\delta x} cos x\)

Back to the proofs:

Property B

Geometric proof:

Lets rename \(\Delta x\) as \(\theta\).

Consider the ratio of the arc length to the line \(sin \theta\). Then double it for the two \(\theta\) that were drawn:

\[
\begin{align}
&\frac{2 sin \theta}{2 \theta} \\
= &\frac{sin \theta}{\theta}
\end{align}
\]

Now, as \(\theta\) approaches 0, the lines \(sin \theta\) and \(arclength \theta\) get close and close to the same line, so the result of the ratio is 1, intuitively because short pieces of curves are nearly straight.

Property A

The basis for the proof for cosine is the same as the proof for sine – that of short pieces of curves being nearly straight.

First, \(cos \theta\) is going to be negative, so we need to invert the statement:

\[
\frac{1 – cos \theta}{\theta}
\]

This is a similar image as before, only magnified.
The distance from the centre of the circle to the circumference is 1 (because this is a unit circle, and the gap between the \(sin \theta\) line and the circumference is \(1 – cos \theta\). As \(\theta\) tends to 0, this gap gets smaller and smaller, and as such, \(\frac{1 – cos \theta}{\theta}\) tends to 0.

NOTE TO SELF – RECORD GEOMETRIC PROOF

General Derivative Rules

Product Rule

\[
(uv)’ = u’v + uv’
\]

can be remembered as changing one variable at a time. Will be proven as such.

Example:

\[
\frac{\delta}{\delta x} (x^n sin\;x) = nx^{n-1}sin\;x + x^n cos\;x
\]

Proof:

Going back to first principles of calculus, we can rewrite the sum in terms of the rate of change:

\[
\begin{align}
\frac{\delta}{\delta x} uv &= \lim_{\Delta x \to 0} \frac{u(x + \Delta x)v(x + \Delta x) – u(x)v(x)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{u(x + \Delta x)(v(x + \Delta x) – v(x)) + v(x)(u(x + \Delta x) – u(x))}{\Delta x} \qquad \text{(1)} \\
&= \lim_{\Delta x \to 0} u(x + \Delta x)\frac{\delta v}{\delta x} + v(x)\frac{\delta u}{\delta x} \\
&= uv’ + vu’
\end{align}
\]

We converted the sum in (1) to a new form to aid in the proof. You can multiply out (1) to show that it is indeed equal:

\[
\begin{align}
u(x + \Delta x)(v(x + \Delta x) – v(x)) + v(x)(u(x + \Delta x) – u(x)) &= u(x + \Delta x)v(x + \Delta x) – v(x)u(x + \Delta x) + v(x)u(x + \Delta x) – v(x)u(x) \\
&= u(x + \Delta x)v(x + \Delta x) – v(x)u(x)
\end{align}
\]

As \(\Delta x\) tends to 0, then \(u(x + \Delta x)\) approaches \(u(x)\), which will complete the proof.

Quotient Rule

\[
\left(\frac{u}{v}\right)’ = \frac{u’v – uv’}{v^2}
\]

only works when \(v \ne 0\).

Proof:

First, recall that \(\frac{\delta}{\delta x}\) is equal to \(\frac{\Delta y}{\Delta x}\) and we can compute \(\Delta y\) in a separate step, and then just divide by \(\Delta x\).

\[
\begin{align}
\Delta \left( \frac{u}{v} \right) &= \frac{u + \Delta u}{v + \Delta v} – \frac{u}{v} \\
&= \frac{uv + v(\Delta u) – uv – u(\Delta v)}{(v + \Delta v)v} \\
&= \frac{v(\Delta u) – u(\Delta v)}{(v + \Delta v)v} \\
\frac{\Delta \left( \frac{u}{v} \right) }{\Delta x} &= \frac{ \frac{\Delta u}{\Delta x}v – u\frac{\Delta v}{\Delta x} }{(v + \Delta v)v} \\
\frac{\delta}{\delta x} \left( \frac{u}{v} \right) &= \lim_{\Delta x \to 0} \frac{ \frac{\delta u}{\delta x}v – u\frac{\delta v}{\delta x} }{v^2} \\
&= \frac{u’v – uv’}{v^2}
\end{align}
\]

Let us reconsider the above, setting \(u = 1\):

\[
\begin{align}
\frac{\delta}{\delta x} \left( \frac{1}{v} \right) &= \frac{-1v’}{v^2} \\
&= -v^{-2}v’
\end{align}
\]

and now, lets consider the above example generally, with \(\frac{1}{x^n}\) represented as \(x^{-n}\):

\[
\begin{align}
\frac{\delta}{\delta x} x^{-n} &= \frac{\delta}{\delta x} \left( \frac{1}{x^n} \right) \\
&= -x^{-2n}nx^{n-1} \\
&= -nx^{-n-1}
\end{align}
\]

You’ll notice that this follows the same pattern that applies to \(x^n\).

Chain rule

We’ll start by derriving the rule, then writing it down.

Proof:

\[
\begin{align}
\frac{\Delta y}{\Delta t} &= \frac{\Delta y}{\Delta x} \frac{\Delta x}{\Delta t} \\
\frac{\delta y}{\delta t} &= \frac{\delta y}{\delta x} \frac{\delta x}{\delta t}
\end{align}
\]

TODO: FIND A BETTER PROOF OF THIS, OR SHOW A MORE EXPLICIT EXAMPLE *

Example:

\[
\begin{align}
\frac{\delta}{\delta t} (sin\;t)^{10} &= x^{10} & \text{Substituting x = sin t} \\
&= 10x^9 cos\;t \\
&= 10(sin\;t)^9 cos\;t & \text{Substituting the value of x back in}
\end{align}
\]

Example 2:

\[
\begin{align}
\frac{\delta}{\delta t} sin(10t) &= sin(x) & \text{Substituting x = 10t} \\
&= cos(x) \times 10 \\
&= cos(10t) \times 10 & \text{Substituting 10t = x back in}
\end{align}
\]

Higher derrivatives *

Refers to multiple steps of differentiation.

Consider the function \(u = u(x)\), then \(u’\) is a new function in and of itself. We can differentiate this function: \(u’’\). This is the second derivative, and it too is a new function, and can be differentiated.

You can continue this to any number of higher derrivatives, second, third, fourth, etc.

Notation:
\[
\begin{align}
u’ &= \frac{\delta u}{\delta x} = \frac{\delta}{\delta x} u \\
u’’ &= \frac{\delta}{\delta x}\frac{\delta u}{\delta x} = \frac{\delta}{\delta x}\frac{\delta}{\delta x} u \\
&= \left(\frac{\delta}{\delta x}\right)^2 u \\
&= \frac{\delta^2}{(\delta x)^2} u \\
&= \frac{\delta^2}{\delta x^2} u
\end{align}
\]

Example:

\[
\begin{align}
D^nx^n &= ? \\
Dx^n &= nx^{n-1} \\
D^2x^n &= n(n-1)x^{n-2} \\
D^3x^n &= n(n-1)(n-2)x^{n-3} \\
D^{n-1}x^n &= (n(n-1)(n-2) \ldots 2)x^1 \\
D^nx^n &= (n(n-1)(n-2) \ldots 2 \times 1) \times 1 \\
&= n!
\end{align}
\]