Exponentials and Logarithms

Goal: Calculate \(\frac{d}{dx} a^x\)

-\(\lim\limits_{\Delta x\rightarrow0} \frac{a^{x+\Delta x} - a^x}{\Delta x}\)

As \(a^{x + \Delta x} = a^xa^{\Delta x}\), \(\lim\limits_{\Delta x\rightarrow0} a^x \frac{a^{\Delta x} - 1}{\Delta x}\)

\(a^x\) is a constant so \(a^x \lim\limits_{\Delta x\rightarrow0} \frac{a^{\Delta x} - 1}{\Delta x}\)

DEFN \(M(a) := \lim\limits_{\Delta x\rightarrow0} \frac{a^{\Delta x} - 1}{\Delta x}\)

With new definition \(\frac{d}{dx} a^x = M(a)a^x\).
Plug in \(x=0\) to get \(\frac{d}{dx} a^0 = M(a)\), showing \(M(a)\) is the slope at 0.

What is M(a)?

• Define base \(e\) as the unique number such that \(M(e) = 1\)
  
• If this is the case \(\frac{d}{dx} e^x = e^x\) (as \(\frac{d}{dx} a^x = M(a)a^x\)).
  

Why does e exist?

• Take example \(f(x) = 2\), \(f'(0) = M(2)\) and stretch by constant \(k\).
  

\(f(kx) = 2^{kx} = (2^k)^x = b^x\), where \(b = 2^k\).

• As \(k\) is increased the slope of the function gets steeper. \(\frac{d}{dx} b^x = kf'(kx)\)
  
  • At 0, \(\frac{d}{dx} b^x = kf'(kx) = kf'(0) = kM(2)\) so \(b=e\) when \(k = \frac{1}{M(2)}\)
    

Recall that \(\ln{x_1 x_2} = \ln{x_1} + \ln{x_2}\) and \(\ln{1} = 0\) and \(\ln{e} = 1\).

Differentiate \(w = \ln{x}\) implicitly in the form \(e^w = x\):

• \(\frac{d}{dx} e^w = \frac{d}{dx} x = 1\)
  
• \(\frac{d}{dw} e^w \frac{dw}{dx} = 1\) or \(e^w \frac{dw}{dx} = 1\)
  
• Algebra yields \(\frac{dw}{dx} = \frac{1}{e^w} = \frac{1}{x}\)
  

Method 1
Use base \(e\) = \((e^{\ln{a}})^x = e^{x\ln{a}}\).

Just as the derivative of \(e^{3x}\) is \(3e^{3x}\) by chain rule, \(\frac{d}{dx} e^{x\ln{a}} = (\ln{a}) e^{x\ln{a}}}\). So, \(\frac{d}{dx} a^x = (\ln{a})a^x\).

NOTE: No matter what our base (2 or 10 or something else) the derivative is still \((\ln{a})a^x\) and that's one reason why it's the "natural" log as it comes up naturally.

Method 2

HEY I NEED REVISITING REWATCH THE LAST 15 MIN OF THIS

Logarithmic differentiation. Chain rule + differentiation of logarithm.
\((\ln{u})' = \frac{u'}{u}\)

EXAMPLE

\(v = x^x\)
\(\ln{v} = x\ln{x}\)
\((\ln{v})' = \ln{x} + x \frac{1}{x}\)
\(\frac{v'}{v} = 1 + \ln{x}\)
\(v' = v(1+\ln{x})\)
\(\frac{d}{dx} x^x = x^x(1+\ln{x})\)