Why is the derivative of e^x equal to e^x?

Every exponential b^x has derivative b^x times a constant that depends on b. The number e is chosen so that this constant is exactly 1.

What is the derivative of ln(2x)?

frac1x. By the chain rule it is (2)/(2x)=frac1x; or use log rules: ln(2x)=ln2+ln x, and ln2 is a constant.

What is the derivative of 2^x?

2^xln2, about 0.693·2^x. Don't use the power rule: the variable is in the exponent.

When should I use logarithmic differentiation?

When the variable is in both the base and the exponent, like x^x or x^sin x, or for a long product and quotient of many factors.

Derivative of e^x and ln x: Exponential and Log Rules

Q: What is the derivative of e^(-x)?

-e^-x, by the chain rule: the inside -x has derivative -1.

Q: What is the derivative of ln|x|?

frac1x for every xne0. For negative x, ln|x|=ln(-x), and the chain rule gives (-1)/(-x)=frac1x.

What is the derivative of e^x?

The derivative of $e^{x}$ is $e^{x}$ . To see why the number $e \approx 2.718$ is special, look at any exponential $b^{x}$ (a fixed base $b$ raised to the variable power $x$ ). Its derivative from the definition is

\begin{aligned} \frac{d}{d x} b^{x} & = lim_{h \to 0} \frac{b^{x + h} - b^{x}}{h} \\ = b^{x} \cdot lim_{h \to 0} \frac{b^{h} - 1}{h}, \end{aligned}

because $b^{x + h} = b^{x} b^{h}$ , so $b^{x}$ factors out. The remaining limit is a number that depends only on the base: about $0.693$ for $b = 2$ and about $1.099$ for $b = 3$ . The number $e$ is defined as the base for which this limit is exactly $1$ . So for $e^{x}$ the derivative is $e^{x} \cdot 1$ : the function is its own derivative.

Derivative of the natural exponential

\frac{d}{d x} e^{x} =

e^{x}

On a graph, this says that at every point of $y = e^{x}$ , the slope equals the height. At $x = 0$ the height is $1$ , so the tangent line there has slope $1$ .

y = e^{x}

with its tangent line of slope

1

(0, 1)

, and

y = \ln x

, the mirror image of

e^{x}

across the line

y = x

What is the derivative of ln x?

The natural logarithm $\ln x$ is the logarithm with base $e$ : $\ln x$ is the power you raise $e$ to in order to get $x$ . So $y = \ln x$ means $e^{y} = x$ . Differentiating both sides of $e^{y} = x$ with respect to $x$ (using implicit differentiation) gives $e^{y} \frac{d y}{d x} = 1$ , so $\frac{d y}{d x} =$ $\frac{1}{e^{y}} =$ $\frac{1}{x}$ .

Derivative of the natural logarithm

\frac{d}{d x} \ln x =

\frac{1}{x}

for

x > 0

, and more generally

\frac{d}{d x} \ln | x | =

\frac{1}{x}

for

x \neq 0

How do you differentiate other bases, like 2^x and log base 2?

Rewrite the base in terms of $e$ : since $b = e^{\ln b}$ , we have $b^{x} = e^{x \ln b}$ , and the chain rule gives $e^{x \ln b} \cdot \ln b =$ $b^{x} \ln b$ . For logarithms, the change-of-base rule $\log_{b} x =$ $\frac{\ln x}{\ln b}$ turns every log into a natural log divided by a constant.

Function	Derivative	Example
$e^{x}$	$e^{x}$	$\frac{d}{d x} 5 e^{x} =$ $5 e^{x}$
$b^{x}$	$b^{x} \ln b$	$\frac{d}{d x} 2^{x} =$ $2^{x} \ln 2$
$\ln x$	$\frac{1}{x}$	$\frac{d}{d x} 4 \ln x =$ $\frac{4}{x}$
$\log_{b} x$	$\frac{1}{x \ln b}$	$\frac{d}{d x} \log_{10} x =$ $\frac{1}{x \ln 10}$
$e^{g (x)}$	$e^{g (x)} g^{'} (x)$	$\frac{d}{d x} e^{3 x} =$ $3 e^{3 x}$
$\ln g (x)$	$\frac{g^{'} (x)}{g (x)}$	$\frac{d}{d x} \ln (x^{2} + 1) =$ $\frac{2 x}{x^{2} + 1}$

Common mistake

Using the power rule on an exponential. In $x^{n}$ the variable is in the base; in $b^{x}$ it's in the exponent. $\frac{d}{d x} e^{2 x}$ is not $2 x e^{2 x - 1}$ ; it is $2 e^{2 x}$ , by the chain rule.

How do you use the chain rule with e and ln?

Most exponential and log derivatives in practice have an inside function, so they need the chain rule. For $e^{g (x)}$ , the outside $e^{u}$ is its own derivative, so the result is the same exponential times $g^{'} (x)$ . For $\ln g (x)$ , the outside $\ln u$ has derivative $\frac{1}{u}$ , so the result is $\frac{1}{g (x)}$ times $g^{'} (x)$ .

Worked example: An exponential with an inside

Differentiate $y = e^{3 x^{2}}$ .

Inside: $3 x^{2}$ , with derivative $6 x$ . Outside: $e^{u}$ , its own derivative.
Outside's derivative, inside unchanged: $e^{3 x^{2}}$ .
Times the inside's derivative: $y^{'} = 6 x e^{3 x^{2}}$ .

Answer: $y^{'} = 6 x e^{3 x^{2}}$

Worked example: A logarithm with an inside

Differentiate $y = \ln (x^{2} + 1)$ .

Inside: $x^{2} + 1$ , with derivative $2 x$ .
$\frac{d}{d x} \ln g =$ $\frac{g^{'}}{g}$ , so $y^{'} =$ $\frac{2 x}{x^{2} + 1}$ .

Answer: $y^{'} =$ $\frac{2 x}{x^{2} + 1}$

Worked example: A product with a logarithm

Differentiate $y = x^{2} \ln x$ .

Product rule with $f = x^{2}$ ( $f^{'} = 2 x$ ) and $g = \ln x$ ( $g^{'} = \frac{1}{x}$ ).
$y^{'} =$ $2 x \ln x + x^{2} \cdot \frac{1}{x}$ .
Simplify $x^{2} \cdot \frac{1}{x} = x$ : $y^{'} = 2 x \ln x + x$ , which factors as $x (2 \ln x + 1)$ .

Answer: $y^{'} = x (2 \ln x + 1)$

Should you use log rules before differentiating?

Yes, whenever a logarithm contains a product, a quotient or a power. The rules $\ln (A B) = \ln A + \ln B$ , $\ln \frac{A}{B} =$ $\ln A - \ln B$ and $\ln (A^{k}) = k \ln A$ turn one hard derivative into several easy ones.

Worked example: Log rules first

Differentiate $y =$ $\ln (x^{3} \sqrt{x + 1})$ .

Write the root as a power: $\sqrt{x + 1} =$ $(x + 1)^{1 / 2}$ .
Split the product, then bring the powers down: $y =$ $\ln x^{3} + \ln (x + 1)^{1 / 2} =$ $3 \ln x + \frac{1}{2} \ln (x + 1)$ .
Differentiate each term: $y^{'} =$ $\frac{3}{x} + \frac{1}{2} \cdot \frac{1}{x + 1}$ .
Tidy: $y^{'} =$ $\frac{3}{x} + \frac{1}{2 (x + 1)}$ .

Answer: $y^{'} =$ $\frac{3}{x} + \frac{1}{2 (x + 1)}$

Common mistake

Splitting a log of a sum. There is no rule for $\ln (A + B)$ ; in particular $\ln (x^{2} + 1) \neq \ln x^{2} + \ln 1$ . Log rules only split products, quotients and powers.

What is logarithmic differentiation?

Some functions fit none of the rules directly, most famously $x^{x}$ , which has the variable in both the base and the exponent (so neither the power rule nor the exponential rule applies). Logarithmic differentiation takes $\ln$ of both sides, uses log rules to bring the exponent down, and then differentiates implicitly. It is also a shortcut for long products and quotients of many factors.

Worked example: A variable base and exponent

Differentiate $y = x^{x}$ for $x > 0$ .

Take $\ln$ of both sides: $\ln y =$ $\ln (x^{x}) =$ $x \ln x$ .
Differentiate both sides with respect to $x$ . Left: $\frac{1}{y} \frac{d y}{d x}$ (chain rule). Right, by the product rule: $1 \cdot \ln x + x \cdot \frac{1}{x} =$ $\ln x + 1$ .
So $\frac{1}{y} \frac{d y}{d x} =$ $\ln x + 1$ . Multiply both sides by $y$ : $\frac{d y}{d x} =$ $y (\ln x + 1)$ .
Replace $y$ by $x^{x}$ : $\frac{d y}{d x} =$ $x^{x} (\ln x + 1)$ .

Answer: $\frac{d y}{d x} =$ $x^{x} (\ln x + 1)$

Why do exponential functions describe growth and decay?

Because $\frac{d}{d t} C e^{k t} =$ $k \cdot C e^{k t}$ , the function $P (t) = C e^{k t}$ has a rate of change proportional to its own size: $P^{'} = k P$ . That's exactly how populations, investments with continuous interest, and radioactive material behave: the more there is, the faster it changes. A positive $k$ gives growth and a negative $k$ gives decay.

Derivatives of eˣ, ln x, and other exponential and log functions