Why Small Errors Change the Conclusion
A derivative from first principles is defined by a limit. That means a correct calculation depends on both the algebra in the difference quotient and the behavior of that quotient as the input approaches the point. A single careless substitution or cancellation can change a valid limit into an undefined expression, or make a nonexistent derivative appear to exist.
We will focus on four recurring errors: setting the increment equal to zero too early, canceling without factoring, replacing a function value with a limit, and checking only one side when both sides are required.
The quotient is evaluated for nonzero \(h\), then its limit is considered as \(h\) approaches zero. The expression at \(h=0\) is not itself the derivative: direct substitution there gives a zero denominator.
Worked Example: Do Not Substitute the Base Point Too Soon
Worked Example: Finding the derivative of a square at \(a=3\)
Let \(f(x)=x^2\). Then \(f(3)=3^2=9\). By the definition, for \(h\ne 0\), $$ \frac{f(3+h)-f(3)}{h} = \frac{(3+h)^2-9}{h}. $$ Expand the square: $$ \frac{(3+h)^2-9}{h} = \frac{9+6h+h^2-9}{h} = \frac{6h+h^2}{h}. $$ Factor \(h\) in the numerator and cancel the common factor, which is permitted because \(h\ne0\): $$ \frac{h(6+h)}{h}=6+h. $$ Now take the limit: $$ f'(3)=\lim_{h\to0}(6+h)=6. $$ Check: the simplified quotient is exactly \(6+h\) for every nonzero \(h\), and its limit as \(h\to0\) is \(6\). Substituting \(h=0\) into the original quotient would give \(0/0\), which is undefined; it is not a step in evaluating the limit.
The order matters: simplify the quotient on its allowed inputs first, then find its limit. The fact that the limiting increment is zero does not mean the increment can be set to zero inside the original quotient.
Why Simplifying Before Taking the Limit Works
A limit describes values arbitrarily close to the point, not necessarily the value of the expression at the point. Consequently, two expressions that agree for all sufficiently small nonzero \(h\) have the same limit as \(h\to0\), if either limit exists. Here is the reason directly from the definition of a limit.
This is why factoring and canceling a nonzero increment can be useful: the simplified expression agrees with the original difference quotient at every permitted nonzero increment near zero. It does not justify canceling terms that are not common factors.
Cancel Factors, Not Pieces of a Sum
Cancellation is valid when the numerator and denominator contain the same nonzero multiplicative factor. It is not an operation that can be applied to just one part of a sum while leaving the other part unchanged. When a difference quotient contains several terms, first factor the numerator to reveal any common factor.
Worked Example: Factor before canceling
Let \(f(x)=x^2+2x\) and \(a=1\). First calculate the base-point value: $$ f(1)=1^2+2(1)=1+2=3. $$ For \(h\ne0\), the difference quotient is $$ \frac{f(1+h)-f(1)}{h} = \frac{(1+h)^2+2(1+h)-3}{h}. $$ Expand each term: $$ \frac{1+2h+h^2+2+2h-3}{h} = \frac{4h+h^2}{h}. $$ Factor the numerator: $$ \frac{h(4+h)}{h}=4+h, $$ where cancellation is valid because \(h\ne0\). Therefore, $$ f'(1)=\lim_{h\to0}(4+h)=4. $$ Check: multiplying the simplified result \(4+h\) by the denominator \(h\) gives \(h(4+h)=4h+h^2\), the original numerator. A tempting but invalid cancellation would turn \((4h+h^2)/h\) into \(4+h^2\) by canceling the denominator only against \(4h\). That result fails the multiplication check: \(h(4+h^2)=4h+h^3\), not \(4h+h^2\).
The check at the end is useful: after simplifying a fraction, multiply the proposed quotient by its denominator. The product must recover the original numerator wherever the denominator is nonzero.
Use the Actual Function Value at the Base Point
In the definition, the second function value in the numerator is \(f(a)\), the value assigned to the function at the point. It is not automatically the limit of \(f(x)\) as \(x\to a\). Those two numbers are equal when \(f\) is continuous at \(a\), but continuity must be known; it cannot be assumed just because the limit exists.
Worked Example: A limit is not the assigned function value
Define $$ f(x)= \begin{cases} x^2,&x\ne0,\\ 5,&x=0. \end{cases} $$ The values \(f(h)\) approach \(0\) as \(h\to0\), but the actual base-point value is \(f(0)=5\). Thus, for \(h\ne0\), the derivative quotient at zero is $$ \frac{f(h)-f(0)}{h} = \frac{h^2-5}{h} = h-\frac{5}{h}. $$ As \(h\to0^+\), \(h\to0\) and \(5/h\to+\infty\), so \(h-5/h\to-\infty\). As \(h\to0^-\), \(5/h\to-\infty\), so \(h-5/h\to+\infty\). The two-sided limit is not finite; therefore \(f'(0)\) does not exist. Check: for every nonzero \(h\), the numerator uses \(f(0)=5\), exactly as the definition requires. If one wrongly replaced \(f(0)\) by the limit value \(0\), the quotient would instead be \(h^2/h=h\), leading to the false conclusion that the derivative is \(0\).
This example also shows why a function’s limit near a point cannot repair a different value assigned at the point. The derivative quotient measures changes relative to the actual value \(f(a)\).
Check Both Sides at an Interior Point
For an interior point, the two-sided limit in the derivative definition requires agreement from both directions. The right-hand and left-hand difference-quotient limits must both exist as finite numbers and be equal. A calculation from only one side can establish a one-sided derivative, but it cannot establish the ordinary two-sided derivative at an interior point.
Worked Example: The two sides disagree for an absolute value
Let \(f(x)=|x|\) and consider \(a=0\). Since \(f(0)=0\), the quotient for \(h\ne0\) is $$ \frac{f(0+h)-f(0)}{h} = \frac{|h|}{h}. $$ For \(h>0\), \(|h|=h\), so the quotient is \(h/h=1\). Hence the right-hand limit is \(1\). For \(h<0\), \(|h|=-h\), so the quotient is \((-h)/h=-1\). Hence the left-hand limit is \(-1\). Since \(1\ne-1\), the two-sided limit does not exist and \(f'(0)\) does not exist. Check: the quotient is exactly \(1\) on all negative-increment-free inputs \(h>0\), and exactly \(-1\) for \(h<0\); the two one-sided limits are therefore as stated. Checking only \(h>0\) would give the right-hand derivative \(1\), not a two-sided derivative.
At a domain endpoint, there may be no inputs on one side. In that case, the relevant derivative convention is one-sided: only increments that keep \(a+h\) in the domain are used. At an interior point, however, overlooking one side can conceal a failure of differentiability.
A Reliable First-Principles Checklist
Before accepting a derivative calculation, check each stage in order. These checks do not replace the limit argument; they help ensure the argument uses the definition correctly.
Use \(\bigl(f(a+h)-f(a)\bigr)/h\), and keep \(h\ne0\) while simplifying.
Calculate \(f(a)\) from the function’s definition, including any special assigned value at \(a\).
Expand or factor as needed. Cancel only a common multiplicative factor that is nonzero on the quotient’s domain.
At an interior point, check both sides; at an endpoint, use the permitted one-sided limit.
Undo a cancellation by multiplication, and confirm that the required limit is finite and, when two-sided, that both sides agree.
Why These Checks Matter
First-principles work is especially sensitive to the distinction between an expression and its limit. The original quotient is undefined at \(h=0\), yet it can still have a finite limit there after valid simplification. Conversely, a simplified expression obtained through invalid cancellation may have a neat limit that says nothing about the original quotient.
Likewise, the derivative definition uses both the function’s assigned value and nearby function values. Replacing either with a different quantity changes the quotient. Finally, the direction of approach is part of the limit: at an interior point both sides matter, while at an endpoint the domain determines which side is available. Keeping these requirements visible makes first-principles calculations more reliable and helps distinguish a correct derivative from a plausible-looking but unsupported answer.
Check Your Understanding
For each question, identify the condition or step that determines the correct conclusion.
- Why is it invalid to substitute \(h=0\) into the original difference quotient before simplifying?
- For what reason is canceling a common factor \(h\) allowed only after noting \(h\ne0\)?
- What is the difference between \(f(a)\) and \(\lim_{x\to a}f(x)\), and when must they be equal?
- If the right-hand difference-quotient limit is \(2\) and the left-hand limit is \(-2\), does the derivative exist at an interior point? Explain.
- When \(a\) is an endpoint of the domain, which increments may be used in the derivative limit?
