From Secant Slopes to a Derivative
A secant line passes through two points on the graph of a function. If one point is fixed and the other moves toward it, the slope of the secant line may approach a single value. That limiting value is the derivative at the fixed point.
This description involves two different roles. The point where we want the derivative is held fixed. A separate variable moves toward that point. Keeping those roles distinct is essential: the derivative is defined by a limit of slopes, not by substituting the point directly into a quotient that is undefined there.
An accumulation point of \(A\) is a point that can be approached by points of \(A\) other than itself. This condition ensures that there are nearby values of \(x\) at which the quotient can be evaluated. When \(A\) contains an interval on both sides of \(a\), the limit requires the quotient to approach the same finite value from both sides.
What the Difference Quotient Measures
For \(x\ne a\), the graph contains the points \((a,f(a))\) and \((x,f(x))\). The slope of the line through these points is
The numerator is the change in function value, and the denominator is the change in input. Their ratio is the average rate of change between the two inputs, or equivalently the secant slope. As \(x\) approaches \(a\), the second point moves toward the fixed point on the graph. The derivative, when it exists, is the finite value approached by those secant slopes.
The variable \(x\) is the quantity that changes in the limit. The number \(a\) remains fixed throughout. Accordingly, \(f(a)\) is also fixed, while \(f(x)\) varies with \(x\). The notation \(f'(a)\) names the resulting derivative value at the point \(a\); it does not mean that \(a\) is being varied inside this particular limit.
What It Means for the Limit to Exist
Saying that the derivative limit exists means more than saying that some sample secant slopes look close to one another. They must approach one finite number as the input gets arbitrarily close to \(a\), from every direction allowed by the domain.
In words, no matter how small a tolerance \(\varepsilon\) is chosen for the slopes, all sufficiently nearby secant slopes must lie within that tolerance of \(L\). The condition \(0<|x-a|\) excludes \(x=a\), where the denominator would be zero. The limit concerns nearby inputs, not the quotient evaluated at the point itself.
The word finite matters. Secant slopes that grow without bound do not approach a real number and therefore do not define a derivative under this definition. Nor is it enough for the left-hand and right-hand slopes to approach different finite values: the two-sided limit then fails to exist.
Worked Examples
Worked Example: A Quadratic at a Fixed Point
Find the derivative of \(f(x)=x^2\) at \(a=3\) from the definition.
First identify the fixed point and its function value:
For \(x\ne3\), the difference quotient is
The cancellation is valid because the quotient is being considered only when \(x\ne3\). Now take the limit:
To check the limit precisely, let \(\varepsilon>0\) and choose \(\delta=\varepsilon\). If \(0<|x-3|<\delta\), then
Thus the quotient approaches \(6\), as claimed. For a numerical check, at \(x=3.1\) the secant slope is \(3.1+3=6.1\), and at \(x=2.9\) it is \(2.9+3=5.9\). These values approach \(6\) from opposite sides as \(x\) approaches \(3\).
Worked Example: A Constant Function
Let \(f(x)=7\) for every real \(x\). Find the derivative at \(a=4\).
The fixed function value is \(f(4)=7\). For any \(x\ne4\),
The quotient is exactly \(0\) for every allowed \(x\), so its limit as \(x\to4\) is \(0\):
For a check using the limit condition, let \(\varepsilon>0\) and choose any \(\delta>0\), for example \(\delta=1\). Whenever \(0<|x-4|<\delta\), the difference between the quotient and \(0\) is \(|0-0|=0<\varepsilon\). Therefore the derivative exists and equals \(0\).
Worked Example: Secant Slopes Do Not Approach a Finite Value
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(0)=1\) and \(f(x)=0\) for \(x\ne0\). Determine whether \(f'(0)\) exists.
The fixed point is \(a=0\), and \(f(0)=1\). For every \(x\ne0\), \(f(x)=0\), so
As \(x\) approaches \(0\) from the positive side, \(-1/x\) decreases without bound. From the negative side, it increases without bound. In particular, the quotient cannot approach a finite real number.
For a direct check, suppose it had a finite limit \(L\). The limit condition with \(\varepsilon=1\) would give some \(\delta>0\) such that the quotient is within \(1\) of \(L\) whenever \(0<|x|<\delta\). Choose a positive \(x\) satisfying
Then \(1/x>|L|+2\), and the triangle inequality gives
This contradicts the required condition. Hence there is no finite limit, and \(f'(0)\) does not exist.
Why the Definition Works
The definition translates the geometric idea of a tangent slope into a precise limiting statement. At every \(x\ne a\), the quotient gives an actual secant slope, which can be computed from two distinct points. If all such slopes sufficiently close to \(a\) are close to one finite number \(L\), then \(L\) is the slope approached by the secants as their moving point reaches the fixed point.
This explains why a derivative is not simply “the slope between the points at \(a\) and \(a\).” Those are the same point, so that expression would have zero change in both the input and the function value and would not define a secant slope. Instead, the derivative records what the slopes between distinct nearby points approach.
The limit condition also makes the role of existence precise. A few slopes being close to \(L\) is not enough; the condition must hold for every sufficiently close input, to any requested tolerance. If different approaches produce incompatible values, or the slopes become unbounded, no single finite derivative value satisfies the definition.
A Common Mistake: Substituting Too Soon
A frequent error is to set \(x=a\) in the difference quotient before taking the limit. For the quadratic example at \(a=3\), this gives the incorrect calculation
The conclusion does not follow: \(0/0\) is undefined, not equal to \(0\). The quotient is defined for \(x\ne3\), and its limit is found by simplifying it for those \(x\) and then letting \(x\) approach \(3\):
The corrected reasoning never evaluates the original quotient at \(x=3\). It studies its values at nearby inputs, exactly as the definition requires.
Reading the Notation Carefully
The notation \(f'(a)\) identifies the derivative at one specified point. If the point varies, the notation \(f'(x)\) refers to the derivative function, when that function is defined. In either case, the definition at a particular point uses a fixed input and a variable approaching it:
A reliable reading procedure is to locate \(a\) first, calculate \(f(a)\), and then treat \(x\) as the moving input. Ask whether the quotient approaches one finite value through all nearby domain points. If it does, that value is \(f'(a)\); if it does not, the derivative at \(a\) is undefined.
Check Your Understanding
For each question, pay attention to which input is fixed and which is allowed to approach it.
- In the definition of \(f'(a)\), which symbol names the fixed point, and which symbol varies in the limit?
- For \(f(x)=x^2\) at \(a=2\), write the difference quotient without simplifying it.
- What does the difference quotient represent geometrically when \(x\ne a\)?
- State in words what it means for the derivative limit to exist as a finite number.
- Why does substituting \(x=a\) into the difference quotient not calculate the derivative?
- Can a function have a derivative at \(a\) if its secant slopes approach different values from opposite sides? Explain.
