Differentiation Rules

• \(\frac{d}{dx} cu = c \frac{d}{dx} u\)
  
• \(\frac{d}{dx} u + v = \frac{d}{dx} u + \frac{d}{dx} v\)
  

Differentiating a product of functions: rule is \((uv)' = u'v + uv'\)

PROOF

\(\Delta(uv) = u(x+\Delta x)v(x+\Delta x) - u(x)v(x)\)

\(= (u(x+\Delta x) - u(x))v(x + \Delta x) + u(x)v(x+\Delta x) - u(x)\)

\(= (\Delta u)(v(x+\Delta x) + u(x)\Delta v\)

\(\frac{\Delta(uv)}{\Delta x} = \frac{\Delta(u)}{\Delta x} v(x+\Delta x) + u \frac{\Delta v}{\Delta x}\)

Take limit to get \(\frac{d(uv)}{d x} = \frac{du}{dx} v + u \frac{dv}{dx}\)

Differentiating a quotient of functions: rule is \(\big(\frac{u}{v}\big)' = \frac{u'v - uv'}{v^2}\)

PROOF

TIMESTAMP: ~15:00 Lecture 4
Type me out later!

Shows us that Power Rule works for negative powers!

EXAMPLE \(y = sin(x)^{10}\)

Solution is to add intermediate variable names.

EXAMPLE \(u = sin(x)\), \(y = u^{10}\)

*PROOF

*
\(\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u}\frac{\Delta u}{\Delta x}\)

The \(\Delta u\) cancels!

As \(\Delta x\) goes to 0:

\(\frac{dy}{dx} = \frac{dy}{du}\frac{du}{dx}\)

Differentiation of a composition is a product.

EXAMPLE
\(y = sin(x)^{10}\)
Introduce intermediate variable:
\(u = sin(x)\), \(y = u^{10}\)
\(\frac{dy}{du} = cos(x)\)
\(\frac{du}{dy} = 10x^9\)
Multiply to get:
\(\frac{dy}{dx} = 10u^9 cos(x)\)
Substitute \(u\) to end up with:
\(\frac{dy}{dx} = 10(sin(x))^9 cos(x)\)

WARN: Variable names are confusing…

You can skip the intermediate calculations when trying to to calculate it quickly.

Rinse and repeat.

\(u = u(x)\)
\(u'\) is a new function which can be differentiated again to get \(u''\)
EXAMPLE \(u = \sin(x)\), \(u' = \cos(x)\), \(u'' = -\sin(x)\), \(u''' = -\cos(x)\)…

Sometimes notation is $u⁽⁴⁾ instead of \(u''''\).

The other notation, specifically l\(\frac{d}{dx}\), has an "operator" \(d\) which is applied to a function to get another function (that is the derivative). This can be just \(D\) instead of a fraction sometimes.

Lots of notation.
\(u'' = \frac{d}{dx} \frac{du}{dx} = \frac{d}{dx} \frac{d}{dx} u = \big(\frac{d}{dx}\big)^2 u = \frac{d^2}{(dx)^2} u = \frac{d^2u}{dx^2}\)