2.7 Derivatives of Exponential and Logarithmic Functions

Before we start differentiating exponential and logarithmic functions, we need to start with a review of $\ln, \log$, and $e$

Logarithms are the inverses of exponentials, they tell you to what exponent a number must be raised to achieve a certain number.

$\log_{10}(1) = 0$ because $10^0=1$
$\log_{10}(10) = 1$ because $10^1=10$
$\log_{10}(100) = 2$ because $10^2=100$
$\log_{10}(0) = \text{DNE}$ because $10^n$ will never equal 0 with $n\in \mathbb{R}$

Exponentials take an exponent and produce a number. Logarithms take a number and tell you the exponent that produced it.

They simplify exponential expressions by turning multiplicative/exponential growth into addition and subtraction.

Rules of Logarithms:

$\log_{e}(x)=\ln(x)$
$\log_{b}(xy)=\log_{b}(x)+\log_{b}(y)$
$\log_{b}(\frac{x}{y})=\log_{b}(x)-\log_{b}(y)$
$\log_{b}(x^n)=n\log_{b}(x)$

Most often we will use natural log $\ln$ in calculus which is defined as $\log_{e}(x)=\ln(x)$, a logarithm with base $e$.

$\ln(1) = 0$ because $e^0=1$
$\ln(e) = 1$ because $e^1=e$
$\ln(0) = \text{DNE}$ because $e^n$ will never equal 0 with $n\in \mathbb{R}$

So, how do we differentiate these different functions?

Proof of $e^x$

$$\frac{d}{dx}[e^x]$$

With the limit definition of derivative:

$$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

$$\lim_{h\to0}\frac{e^{x+h}-e^x}{h}$$

$$=\lim_{h\to0} e^x\frac{e^h – 1}{h}$$

Remember: $\lim_{h\to0}\frac{e^h – 1}{h} = 1$

Therefore:

$$\frac{d}{dx}[e^x]=e^x$$

We have defined the derivative of $e^x$, which is just itself! This is an important relationship to remember as it is extremely useful in calculus: The slope of the function $e^x$ at any point is simply $e^x$.

Function	Derivative
$$e^x$$	$$\frac{d}{dx}[e^x]=e^x$$
$$\ln(x)$$	$$\frac{d}{dx}[\ln(x)]=\frac{1}{x}$$
$$x^x$$	$$\frac{d}{dx}[a^x]=a^x\ln(x)$$ Where $a \in \mathbb R$
$$\log_a(x)$$	$$$\frac{d}{dx}[\log_a(x)]=\frac{1}{x\ln(a)}$$ Where $a \in \mathbb R$