How to Find a Derivative Using the Difference Quotient

The difference quotient is the limit definition of a derivative. It measures the average rate of change between two nearby inputs, then lets the distance between them shrink to zero. Since no particular function is specified here, the method below shows how to use the quotient for a general function and works through a complete example.

The difference quotient definition

For a function $f$, the derivative at an input $x$ is defined by

$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h},$$

provided this limit exists. The number $h$ is the change in the input. The two function values, $f(x+h)$ and $f(x)$, give the corresponding change in output. Dividing by $h$ gives the average rate of change; taking the limit gives the instantaneous rate of change.

The same definition can find the derivative at one specified point $a$:

$$f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.$$

Use $x$ when you want a formula for the derivative function, and use $a$ when you only need the slope at a particular input.

A step-by-step method

1. Start with the appropriate definition: use $x$ for a derivative formula or the given point $a$ for a derivative value.
2. Substitute the function into both terms in the numerator. In particular, calculate $f(x+h)$ by replacing every $x$ in the function with $x+h$.
3. Expand brackets and simplify the numerator. Factor out $h$ if possible, then cancel it with the denominator. This cancellation is valid for $h\ne 0$, which is exactly the condition in the difference quotient before the limit is taken.
4. Take the limit as $h$ approaches zero. If the simplified expression is defined at $h=0$, direct substitution usually gives the limit.

Worked example: differentiate $f(x)=x^2+3x$

Find the derivative from the definition. First substitute $x+h$ everywhere that $x$ appears:

$$\begin{aligned} f(x+h)&=(x+h)^2+3(x+h)\\ &=x^2+2xh+h^2+3x+3h. \end{aligned}$$

Now subtract $f(x)=x^2+3x$ and divide by $h$:

$$\begin{aligned} \frac{f(x+h)-f(x)}{h} &=\frac{x^2+2xh+h^2+3x+3h-(x^2+3x)}{h}\\ &=\frac{2xh+h^2+3h}{h}\\ &=2x+h+3,\qquad h\ne 0. \end{aligned}$$

The first line requires parentheses around $f(x)$ because the entire value is being subtracted. After canceling the common factor of $h$, take the limit:

$$f'(x)=\lim_{h\to 0}(2x+h+3)=2x+3.$$

For example, the slope at $x=2$ is $f'(2)=2(2)+3=7$. The difference quotient can also find that value directly by setting $x=2$ from the start:

$$\begin{aligned} f'(2)&=\lim_{h\to 0}\frac{f(2+h)-f(2)}{h}\\ &=\lim_{h\to 0}\frac{(2+h)^2+3(2+h)-10}{h}\\ &=\lim_{h\to 0}(7+h)=7. \end{aligned}$$

Common mistakes

• Substituting incorrectly into $f(x+h)$. Replace every occurrence of $x$, including those inside powers and parentheses. For instance, $(x+h)^2$ expands to $x^2+2xh+h^2$, not $x^2+h^2$.
• Forgetting the subtraction parentheses. The numerator is $f(x+h)-f(x)$. Every term in $f(x)$ changes sign when you subtract it.
• Putting $h=0$ too early. The original quotient has denominator zero at $h=0$, so direct substitution there is not allowed. Simplify the quotient for $h\ne 0$ first, then take the limit.
• Treating the quotient as the derivative. Before taking the limit, it is an average rate of change over an interval of width $h$. The derivative is the limit of those values.
• Stopping at a numerical value when a formula is needed. If the task asks for $f'(x)$, keep $x$ variable. Substitute a number only if the task asks for the derivative at a particular point.

When the limit does not exist

A function does not have a derivative at a point if the difference quotient has no finite limit there. For example, for $f(x)=|x|$ at $x=0$, the quotient is

$$\frac{|0+h|-|0|}{h}=\frac{|h|}{h}.$$

This equals $1$ when $h>0$ and $-1$ when $h<0$. The one-sided limits disagree, so $f'(0)$ does not exist. This illustrates why checking the limit—not merely simplifying the quotient—is part of the definition.