Evaluate a derivative from its definition
A quotient that simplifies
For \(f(x) = x^2 - 2x\), the orange secant line from \(x = 3\) to \(x = 3 + h\) has a green rise and a faint run \(h\). Its slope is the difference quotient \(\dfrac{(3+h)^2 - 2(3+h) - 3}{h}\), where 3 is \(f(3)\).
Move the slider to change \(h\). The quotient is 5 at \(h = 1\) and 4.5 at \(h = 0.5\).
To find the pattern, expand the numerator. The constants 9, \(-6\) and \(-3\) cancel, and \(6h - 2h\) leaves \(4h + h^2\).
Every remaining term has \(h\) as a factor, so dividing by \(h\) gives \(4 + h\), and the two orange readouts are equal for every \(h\). As \(h\) approaches 0, \(4 + h\) approaches 4, so \(f'(3) = 4\).
Three stages
To find \(f'(x)\) from the definition, write the quotient with the formula for \(f\), and expand the numerator. In the figure, the dots stand for the terms that remain.
Simplify until \(h\) is a factor of the whole numerator, and cancel it. This is allowed because \(h\) is not 0 in the quotient.
Let \(h\) approach 0. The terms that still contain \(h\) vanish, and what remains is \(f'(x)\).
The result is a formula for \(f'(x)\), so substituting an input gives the slope of the tangent there.
Notation
Write the quotient, the green rise over the faint run \(h\), with the formula for \(f\).
\[\frac{f(x+h)-f(x)}{h}\]Simplify the numerator until every term has the factor \(h\). For a fraction, combine over a common denominator. For a difference of roots, multiply the numerator and the denominator by the conjugate, the same roots joined by a plus sign.
Cancel \(h\). Then let \(h\) approach 0.
\[\cy{f'(x)}=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]The input \(x\) stays fixed while \(h\) changes.
State where the limit exists. This is the domain of \(f'\).
The domain of \(f'\) can be smaller than the domain of \(f\).
When it fails
At the left end of the domain, \(x = -\tfrac12\), the height is 0, so the quotient for \(\sqrt{2x+1}\) is \(\dfrac{\sqrt{2h}}{h}\). It is 2 for \(h = 0.5\), about 14.1 for \(h = 0.01\), and about 141 for \(h = 0.0001\).
As h shrinks, the secant line becomes almost vertical and the quotient grows without bound, so the limit does not exist. A formula for f′ is valid only where the limit exists.
Example 1: a cubic
Use the definition to find \(f'(x)\) for \(f(x) = x^3 - 4x\).
Write the quotient.
\[\frac{(x+h)^3-4(x+h)-(x^3-4x)}{h}\]Expand the cube. The terms \(x^3\) and \(-4x\) cancel with \(-(x^3-4x)\), and what remains is the green rise from \(f(x)\) to \(f(x+h)\).
\[\frac{3x^2h+3xh^2+h^3-4h}{h}\]Divide every term by \(h\).
\[3x^2+3xh+h^2-4\]Let \(h\) approach 0. The terms with \(h\) vanish.
\[\cy{f'(x)=3x^2-4}\]As a check, the tangent is flat where \(f'(x)=0\), at \(x=\pm\frac{2}{\sqrt3}\approx\pm1.15\), and the figure shows \(x = -1.15\).
Example 2: a root
Use the definition to find \(f'(x)\) for \(f(x) = \sqrt{2x+1}\), where \(x > -\tfrac12\).
Write the quotient.
\[\frac{\sqrt{2x+2h+1}-\sqrt{2x+1}}{h}\]A tempting wrong move: Combine the two roots into one root of the difference.
\[\cr{\frac{\sqrt{2h}}{h}}\]Try \(x = 0\) and \(h = 4\). The orange secant from \((0, 1)\) to \((4, 3)\) has slope \(2 \div 4 = 0.5\), but the wrong formula gives the steeper red line, with slope \(\sqrt8 \div 4 \approx 0.71\). A difference of two roots is not the root of the difference.
Multiply the numerator and the denominator by the conjugate. Since \((A-B)(A+B)=A^2-B^2\), the roots in the numerator disappear.
\[\begin{aligned}&\frac{(2x+2h+1)-(2x+1)}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}\\&=\frac{2h}{h(\sqrt{2x+2h+1}+\sqrt{2x+1})}\end{aligned}\]Cancel \(h\). As \(h\) approaches 0, both roots in the denominator approach \(\sqrt{2x+1}\).
\[\cy{f'(x)=\frac{2}{2\sqrt{2x+1}}=\frac{1}{\sqrt{2x+1}}}\]The domain of \(f'\) is \(x>-\tfrac12\). At \(x = 0\), the slope is 1.
Practice
Use the definition to find \(g'(1)\) for \(g(x) = \dfrac{1}{x+2}\).
Combine the two fractions over a common denominator, so that \(h\) becomes a factor of the numerator.
Solution
Write the quotient at \(x = 1\).
\[\frac{\frac{1}{3+h}-\frac13}{h}\]Combine the fractions over the denominator \(3(3+h)\). The numerator becomes \(-h\), and \(h\) cancels.
\[\frac{-h}{3(3+h)\,h}=\frac{-1}{3(3+h)}\]Let \(h\) approach 0, so \(3(3+h)\) approaches 9.
\[\cy{g'(1)=-\frac19}\]
Questions
Questions 1 and 2 use the graph of \(f(x) = \tfrac13x^3 - x\). Question 3 needs a calculation.
The graph has flat tangents at \(x = -1\) and \(x = 1\). Which formula for \(f'(x)\) is zero at both points?
- \(x^2 - 1\)
- \(x^2 + 1\)
- \(x^2 - x\)
Answer
\(x^2 - 1\)
What is the sign of \(f'(x)\) for \(-1 < x < 1\)?
- Negative, because the curve falls and the tangents slope down
- Positive, because the values of \(f\) are positive near \(x = -1\)
- Zero, because the tangents are flat at both ends
Answer
Negative, because the curve falls and the tangents slope down
Use the definition to find \(f'(x)\) for \(f(x) = 2x^2 + 3\).
- \(4x\)
- \(4x + 3\)
- \(4x + 2h\)
- \(2x\)
Answer
\(4x\)
Summary
You write the quotient with the formula for \(f\), expand the numerator, and cancel the factor \(h\).
As \(h\) approaches 0, the terms that still contain \(h\) vanish, and what remains is the slope of the tangent.
This method gives a derivative formula or a single slope when no rule is available, and the result holds only where the limit exists.
This is one of four skills in topic 2.2 of Which & Why: AP Calculus.