Basic Derivative Rules: Power, Sum, Sine, Cosine and e^x

The rules

Working from the definition every time would be slow. A handful of rules, each proved once from the definition, cover most functions:

Rule	Formula
Constant	$\tfrac{d}{dx}\,c = 0$
Power	$\tfrac{d}{dx}\,x^n = n x^{n - 1}$ for any real $n$
Constant multiple	$\tfrac{d}{dx}\,\big(c\,f(x)\big) = c\,f'(x)$
Sum and difference	$\tfrac{d}{dx}\,\big(f(x) \pm g(x)\big) = f'(x) \pm g'(x)$

The last two come straight from the limit laws: a constant factor and a sum both pass through the limit in the definition. Together they mean a polynomial is differentiated term by term.

Why the power rule works

For a positive integer $n$, expand $(x + h)^n$ with the binomial theorem:

$$(x + h)^n = x^n + n x^{n - 1} h + \big(\text{terms with } h^2, h^3, \dots\big)$$

Subtract $x^n$ and divide by $h$. Every term but one still carries a factor of $h$:

$$\frac{(x + h)^n - x^n}{h} = n x^{n - 1} + h\big(\cdots\big) \;\longrightarrow\; n x^{n - 1}$$

Only the term that is linear in $h$ survives the limit, and its coefficient is $n x^{n - 1}$. The rule holds for every real exponent — the previous lesson checked $n = -1$ and $n = \tfrac{1}{2}$ from the definition — with the general proof coming from logarithmic differentiation.

Roots and reciprocals are powers

Before differentiating, rewrite:

$$\sqrt{x} = x^{1/2}, \qquad \sqrt[3]{x^2} = x^{2/3}, \qquad \frac{1}{x^3} = x^{-3}$$

Then the power rule applies directly, and results can be rewritten back: $\tfrac{d}{dx}\sqrt{x} = \tfrac{1}{2}x^{-1/2} = \tfrac{1}{2\sqrt{x}}$.

Sine and cosine

Using $\sin(x + h) = \sin x \cos h + \cos x \sin h$,

$$\frac{\sin(x + h) - \sin x}{h} = \sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}$$

The limit lesson showed $\tfrac{\sin h}{h} \to 1$ and $\tfrac{\cos h - 1}{h} \to 0$. So

$$\frac{d}{dx} \sin x = \cos x, \qquad \frac{d}{dx} \cos x = -\sin x$$

the second by the same method. Both need $x$ in radians.

e^x and ln x

For any base $b > 0$,

$$\frac{b^{x + h} - b^x}{h} = b^x \cdot \frac{b^h - 1}{h}$$

so the derivative of $b^x$ is $b^x$ times the constant $\lim_{h \to 0} \tfrac{b^h - 1}{h}$. That constant is about $0.69$ for $b = 2$ and about $1.10$ for $b = 3$. The number $e \approx 2.71828$ is the base that makes it exactly $1$. Therefore

$$\frac{d}{dx} e^x = e^x, \qquad \frac{d}{dx} \ln x = \frac{1}{x} \quad (x > 0)$$

The rule for $\ln x$ is proved in the lesson on inverse functions.

Worked examples

Common mistakes

Practice problems