What if the function contains a square root?

If $f(x)$ involves a radical, multiply both the numerator and the denominator of the difference quotient by the conjugate of the numerator. This will eliminate the square root in the numerator and allow you to factor out $h$.

How do I handle rational functions?

When $f(x)$ is a fraction, find a common denominator for the terms $f(x+h)$ and $f(x)$ in the numerator. Combine the fractions, simplify, and the $h$ will factor out.

Can I use $\Delta x$ instead of $h$?

Yes. The notation $\Delta x$ is strictly equivalent to $h$. The setup becomes $\lim_{\Delta x o 0} \frac{f(x+\Delta x) - f(x)}{\Delta x}$. The algebraic steps remain exactly the same.

How to find the derivative from the limit definition

The derivative of a function $f(x)$ represents its instantaneous rate of change. It is found by evaluating the limit of the difference quotient as the step size approaches zero.

This method applies to any differentiable function. It is primarily used when explicitly required by a problem statement or when proving the standard derivative rules.

The setup

Write the formal definition of the derivative using the difference quotient: $f'(x) = \lim_{h o 0} \frac{f(x+h) - f(x)}{h}$

The steps

Substitute $x+h$ into the original function to find $f(x+h)$ .
Subtract the original function $f(x)$ from $f(x+h)$ .
Divide the entire expression by $h$ .
Expand and simplify the numerator to factor out an $h$ .
Cancel the $h$ in the numerator with the $h$ in the denominator.
Evaluate the limit by substituting $h = 0$ into the simplified expression.

Checking the result

Apply the standard shortcut rules for differentiation (such as the power rule, product rule, or chain rule) to the original function $f(x)$ . The resulting expression must exactly match the result obtained from the limit definition.

Common errors

Distribution errors: Forgetting to distribute the negative sign to all terms of $f(x)$ during the subtraction step.
Expansion errors: Incorrectly expanding $(x+h)^2$ as $x^2 + h^2$ instead of $x^2 + 2xh + h^2$ .
Premature evaluation: Attempting to plug in $h=0$ before factoring and canceling the $h$ from the denominator, yielding an undefined $0/0$ form.

Worked example

Find the derivative of $f(x) = 3x^2 - 2x$ using the limit definition.

$f'(x) = \lim_{h o 0} \frac{f(x+h) - f(x)}{h}$ $f'(x) = \lim_{h o 0} \frac{[3(x+h)^2 - 2(x+h)] - [3x^2 - 2x]}{h}$ $f'(x) = \lim_{h o 0} \frac{[3(x^2 + 2xh + h^2) - 2x - 2h] - 3x^2 + 2x}{h}$ $f'(x) = \lim_{h o 0} \frac{3x^2 + 6xh + 3h^2 - 2x - 2h - 3x^2 + 2x}{h}$ $f'(x) = \lim_{h o 0} \frac{6xh + 3h^2 - 2h}{h}$ $f'(x) = \lim_{h o 0} \frac{h(6x + 3h - 2)}{h}$ $f'(x) = \lim_{h o 0} (6x + 3h - 2)$ $f'(x) = 6x + 3(0) - 2$ $f'(x) = 6x - 2$

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References: OpenStax Calculus Volume 1, Chapter 3: Derivatives · Stewart Calculus 8th Edition, Chapter 2: Limits and Derivatives