Basic differentiation rules

In the derivative, we found every derivative with a limit. That works, but it takes many steps. In this chapter, we find rules that give derivatives quickly. Each rule comes from the limit definition, so we only have to do the hard work once.

This page covers four rules: the constant rule, the power rule, the constant multiple rule, and the sum and difference rules.

The constant rule

The derivative of a constant function is $0$:

$\dfrac{d}{dx}(c)=0$.

Why: The graph of $f(x)=c$ is a horizontal line. Its slope is $0$ everywhere. With the limit, $\dfrac{f(x+h)-f(x)}{h}=\dfrac{c-c}{h}=0$, so the limit is $0$.

For example, $\dfrac{d}{dx}(7)=0$ and $\dfrac{d}{dx}(\pi)=0$.

The power rule

For any real number $n$,

$\dfrac{d}{dx}(x^n)=nx^{n-1}$.

In words: bring the exponent down in front, and lower the exponent by $1$.

For example, $\dfrac{d}{dx}(x^5)=5x^{5-1}=5x^4$. Also, $\dfrac{d}{dx}(x)=1$, because $x=x^1$ and $1\cdot x^0=1$.

Optional Proof of the power rule (for a positive integer $n$)

Let $f(x)=x^n$. Use the definition of the derivative:

$\begin{align*}&f'(x)\\&=\lim_{h\to 0}\dfrac{(x+h)^n-x^n}{h}\end{align*}$

Expand $(x+h)^n$ with the binomial theorem:

$\begin{align*}&(x+h)^n\\&=x^n+nx^{n-1}h+\dfrac{n(n-1)}{2}x^{n-2}h^2\\&\quad+\cdots+h^n\end{align*}$

Subtract $x^n$. The $x^n$ terms cancel:

$\begin{align*}&(x+h)^n-x^n\\&=nx^{n-1}h+\dfrac{n(n-1)}{2}x^{n-2}h^2\\&\quad+\cdots+h^n\end{align*}$

So, dividing each term by $h$,

$\begin{align*}&f'(x)\\&=\lim_{h\to 0}\Big[nx^{n-1}+\dfrac{n(n-1)}{2}x^{n-2}h\\&\qquad+\cdots+h^{n-1}\Big]\end{align*}$

Let $h$ approach $0$. Every term after the first still has a factor of $h$, so it approaches $0$:

$\begin{align*}&f'(x)=nx^{n-1}\end{align*}$

The power rule also works when $n$ is negative, a fraction, or any real number. The proofs come later in this chapter:

negative integers: the product and quotient rules
fractions: implicit differentiation
any real number: derivatives of exponential and logarithmic functions

Rewrite roots and fractions as powers

To use the power rule, first write the function as a power of $x$. Use these exponent rules (see integer exponents and rational exponents):

$\dfrac{1}{x^n}=x^{-n}$ and $\sqrt[n]{x^m}=x^{m/n}$.

Example 1: Find each derivative.

(a) $\dfrac{d}{dx}(x^7)$ (b) $\dfrac{d}{dx}\left(\dfrac{1}{x^3}\right)$ (c) $\dfrac{d}{dx}\left(\sqrt[3]{x^2}\right)$

Solution:

(a)

$\begin{align*}&\dfrac{d}{dx}(x^7)\\&=7x^{7-1}\\&=7x^6\end{align*}$

(b) Rewrite $\dfrac{1}{x^3}=x^{-3}$. Then

$\begin{align*}&\dfrac{d}{dx}(x^{-3})\\&=-3x^{-3-1}\\&=-3x^{-4}\\&=-\dfrac{3}{x^4}\end{align*}$

$\begin{align*}&\dfrac{d}{dx}(x^{2/3})\\&=\dfrac{2}{3}x^{2/3-1}\\&=\dfrac{2}{3}x^{-1/3}\\&=\dfrac{2}{3\sqrt[3]{x}}\end{align*}$

Example 2: Use the power rule to find the derivatives of $\dfrac{1}{x}$ and $\sqrt{x}$.

Solution:

First, rewrite $\dfrac{1}{x}=x^{-1}$:

$\begin{align*}&\dfrac{d}{dx}\left(\dfrac{1}{x}\right)\\&=\dfrac{d}{dx}(x^{-1})\\&=-1x^{-1-1}\\&=-x^{-2}\\&=-\dfrac{1}{x^2}\end{align*}$

Next, rewrite $\sqrt{x}=x^{1/2}$:

$\begin{align*}&\dfrac{d}{dx}(\sqrt{x})\\&=\dfrac{d}{dx}(x^{1/2})\\&=\dfrac{1}{2}x^{1/2-1}\\&=\dfrac{1}{2}x^{-1/2}\\&=\dfrac{1}{2\sqrt{x}}\end{align*}$

These are the same answers we found with limits in the derivative as a function. The power rule gets there in one step.

The constant multiple rule

If $c$ is a constant and $f$ is differentiable, then

$\dfrac{d}{dx}[c\,f(x)]=c\,f'(x)$.

In words: a constant factor stays in front while you differentiate.

Why: This follows from the constant multiple law for limits.

For example, $\dfrac{d}{dx}(4x^3)=4\cdot 3x^{3-1}=12x^2$.

The sum and difference rules

If $f$ and $g$ are both differentiable, then

$\dfrac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)$

$\dfrac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)$.

In words: differentiate a sum or a difference one term at a time.

Why: These follow from the sum and difference laws for limits.

Together with the power rule, these rules let us differentiate any polynomial.

Example 3: Find $\dfrac{dy}{dx}$ for ${y=4x^5-3x^3+7x-9}$.

Solution:

Differentiate one term at a time:

$\begin{align*}&\dfrac{dy}{dx}\\&=4(5x^{5-1})-3(3x^{3-1})+7(1)-0\\&=4(5x^4)-3(3x^2)+7\\&=20x^4-9x^2+7\end{align*}$

Example 4: Find $f'(x)$ for ${f(x)=\dfrac{3}{x^2}-2\sqrt{x}+5}$.

Solution:

First, rewrite each term as a power of $x$:

$f(x)=3x^{-2}-2x^{1/2}+5$

Now differentiate one term at a time:

$\begin{align*}&f'(x)\\&=3(-2x^{-2-1})-2\left(\tfrac{1}{2}x^{1/2-1}\right)+0\\&=3(-2x^{-3})-2\left(\tfrac{1}{2}x^{-1/2}\right)\\&=-6x^{-3}-x^{-1/2}\\&=-\dfrac{6}{x^3}-\dfrac{1}{\sqrt{x}}\end{align*}$

Example 5: Find $f'(x)$ for $f(x)=(2x+1)^2$.

Solution:

The power rule does not apply directly, because $(2x+1)^2$ is not a power of $x$. So expand first:

$f(x)=4x^2+4x+1$

Then

$\begin{align*}&f'(x)\\&=4(2x^{2-1})+4(1)+0\\&=8x+4\end{align*}$

(In the chain rule, you will learn a way to do this without expanding.)

Tangent lines, horizontal tangents, and velocity

Example 6: Find the equation of the tangent line to $y=x^3-4x$ at $x=1$.

Solution:

The point: $y=1^3-4(1)=-3$. So the point is $(1,-3)$.

The slope: first find the derivative.

$\begin{align*}&y'\\&=3x^{3-1}-4\\&=3x^2-4\end{align*}$

So at $x=1$ the slope is $3(1)^2-4=-1$.

The tangent line:

$\begin{align*}y-(-3)&=-1(x-1)\\y&=-x-2\end{align*}$

Example 7: Find the points on the graph of $y=x^3-3x$ where the tangent line is horizontal.

Solution:

First, find the derivative:

$\begin{align*}&y'\\&=3x^{3-1}-3\\&=3x^2-3\end{align*}$

A horizontal tangent line has slope $0$. So set the derivative equal to $0$:

$\begin{align*}3x^2-3&=0\\x^2&=1\\x&=\pm 1\end{align*}$

At $x=-1$, $y=-1+3=2$. At $x=1$, $y=1-3=-2$. So the tangent line is horizontal at $(-1,2)$ and $(1,-2)$. These are the peak and the low point of the graph in the derivative as a function (Example 5).

Example 8: A ball is dropped, and after $t$ seconds it has fallen $s(t)=4.9t^2$ meters. Find its velocity at any time $t$, and at $t=2$ seconds.

Solution:

Velocity is the derivative of position:

$\begin{align*}&v(t)\\&=s'(t)\\&=4.9(2t^{2-1})\\&=9.8t\ \text{m/s}\end{align*}$

At $t=2$: $v(2)=9.8(2)=19.6$ m/s. This matches the limit we computed in tangent lines and rates of change (Example 4), but in one line.

Summary

Here $c$ is a constant, $n$ is any real number, and $f$ and $g$ are differentiable:

Rule	Formula
Constant	$\dfrac{d}{dx}(c)=0$
Power	$\dfrac{d}{dx}(x^n)=nx^{n-1}$
Constant multiple	$\dfrac{d}{dx}[c\,f(x)]=c\,f'(x)$
Sum	$\dfrac{d}{dx}[f(x)+g(x)]=f'(x)+g'(x)$
Difference	$\dfrac{d}{dx}[f(x)-g(x)]=f'(x)-g'(x)$

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