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Calculus Differentiation

Definition of derivative

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Definition of a Derivative

California Calculus • Standard 4.0 • Differentiation

Definition of a Derivative is the opening topic of Differentiation in the California Calculus Standards. It is aligned to Standard 4.0, which requires students to understand the formal definition of the derivative at a point, its meaning as the slope of the tangent line (Standard 4.1), and as an instantaneous rate of change (Standard 4.2).

The derivative \(f'(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}\) measures how fast a function changes at a point. It is the slope of the tangent line and the instantaneous rate of change, and it underlies the rest of differential calculus.

California Calculus › Differentiation › Definition of a Derivative  —  Standard 4.0

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Theory

The derivative \(f'(a)\) measures how fast a function is changing at \(x=a\). It is defined as a limit of slopes, \(f'(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}\), and it equals both the slope of the tangent line and the instantaneous rate of change at that point.

The derivative of \(f\) at \(x=a\) is the limit of the difference quotient — the slope of the secant line as the two points slide together:

\[f^{\prime}(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}\]

Geometrically, each difference quotient is the slope of a secant line; as \(h\to 0\) those slopes approach the slope of the tangent line. That tangent slope is \(f'(a)\).

The same number is the instantaneous rate of change of \(f\) at \(a\) — for a position function, that rate is velocity.

Key idea: replacing \(a\) with a variable \(x\) gives the derivative function \(f'(x)\), which returns the slope at every point.
A secant line through two points on a curve approaching the tangent line A secant joins the points at a and a plus h; as h shrinks to zero the secant slope approaches the tangent slope, which is the derivative. x y (a, f(a)) (a+h, f(a+h))
Secant slopes approach the tangent slope as \(h\to 0\).
The tangent line at a point, whose slope is the derivative A single tangent line touches the curve at one point; its slope is f prime of a, the instantaneous rate of change there. x y slope = f′(a)
The derivative \(f'(a)\) is the slope of the tangent line at \(a\).

The limit definition (at a point, and as a function):

\[f^{\prime}(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h},\qquad f^{\prime}(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\]

derivative equals the limit of the difference quotient

The tangent line at \(\big(a,f(a)\big)\) uses the point-slope form:

\[y-f(a)=f^{\prime}(a)\,(x-a)\]

tangent line uses point slope form with slope f prime of a

Differentiable: \(f\) is differentiable at \(a\) when this limit exists. A corner, a cusp, or a vertical tangent makes the limit fail, so the function is not differentiable there.

How to differentiate from the definition

  1. Form the difference quotient \(\dfrac{f(x+h)-f(x)}{h}\).
  2. Expand and simplify until every term has a factor of \(h\) to cancel with the bottom.
  3. Take the limit \(h\to 0\) in the simplified expression.
For a tangent line: compute \(f^{\prime}(a)\) for the slope, find the point \((a,f(a))\), then substitute into \(y-f(a)=f'(a)(x-a)\).
Example 1 — From the limit definition
Use \(f'(x)=\lim_{h\to 0}\dfrac{f(x+h)-f(x)}{h}\) to differentiate \(f(x)=x^2\).
Solution

Expand \((x+h)^2\) and simplify before the \(h\to 0\) step.

\(\dfrac{(x+h)^2-x^2}{h}\)\(=\)\(\dfrac{x^2+2xh+h^2-x^2}{h}\)
\(=\)\(\dfrac{2xh+h^2}{h}=2x+h\)
\(\lim_{h\to 0}(2x+h)\)\(=\)\(2x\)

derivative of x squared is 2x

Example 2 — Slope of the tangent line
Find the equation of the tangent to \(y=x^2\) at \(x=3\).
Solution

The slope is \(f'(3)\), and the point is \((3,9)\).

\(f'(x)\)\(=\)\(2x\)
\(f'(3)\)\(=\)\(6\)
\(y-9\)\(=\)\(6(x-3)\)

So the tangent line is \(y=6x-9\).

tangent line y equals 6x minus 9

Example 3 — Instantaneous rate of change
An object has position \(s(t)=t^2\) ft. Find its velocity at \(t=4\) s.
Solution

Velocity is the derivative of position, \(s'(t)\).

\(s'(t)\)\(=\)\(2t\)
\(s'(4)\)\(=\)\(8\)

The velocity is \(8\) ft/s.

velocity equals 8 feet per second

Example 4 — A reciprocal from the definition
Use the limit definition to differentiate \(f(x)=\dfrac{1}{x}\).
Solution

Combine the two fractions over a common denominator first.

\(\dfrac{\dfrac{1}{x+h}-\dfrac{1}{x}}{h}\)\(=\)\(\dfrac{x-(x+h)}{h\,x(x+h)}\)
\(=\)\(\dfrac{-1}{x(x+h)}\)
\(\lim_{h\to 0}\)\(=\)\(-\dfrac{1}{x^2}\)

derivative of one over x is negative one over x squared

Common pitfalls

Never set \(h=0\) too early. Simplify until the \(h\) on the bottom cancels; only then take the limit.
The derivative is a slope, not a value. \(f'(a)\) tells you how fast \(f\) changes at \(a\), which is different from \(f(a)\) itself.
Not every function is differentiable. A sharp corner or a vertical tangent has no well-defined slope, so \(f'\) does not exist there.

Frequently asked questions

What is the definition of a derivative?

The limit of the difference quotient: \(f'(a)=\lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h}\). It is the slope of the tangent line at \(a\).

What does the derivative represent?

The instantaneous rate of change of the function, which equals the slope of the tangent line at that point. For position, it is velocity.

How do you find the equation of a tangent line?

Compute the slope \(f'(a)\), find the point \((a,f(a))\), then use point-slope form \(y-f(a)=f'(a)(x-a)\).

When is a function not differentiable?

Where the difference-quotient limit fails: at a sharp corner, a cusp, a jump, or a vertical tangent. A function must at least be continuous to be differentiable.