Why Start With a Difference Quotient?
A derivative formula is useful only if we know what it says and where it applies. For a function with a variable in the denominator, the denominator also restricts the inputs at which the function—and therefore its derivative—can be defined. Here we will derive the formula for a reciprocal linear function directly from the derivative definition, keeping that restriction visible throughout.
The function considered is \(f(x)=c/(x-a)\), where \(a\) and \(c\) are fixed real constants. The calculation will use the difference quotient at an input \(x\), with an increment \(h\), and then take the limit as \(h\) approaches zero. The main algebraic step is to combine two fractions before taking that limit.
The Domain and the Difference Quotient
The denominator \(x-a\) is zero when \(x=a\), so \(f(a)\) is not defined. Consequently, the derivative at \(a\) cannot be formed from the derivative definition: that definition requires the value \(f(a)\). The formula we derive will apply only when \(x\ne a\).
Fix such an input \(x\). To make \(f(x+h)\) defined as well, we also need \(x+h\ne a\). Thus, in the difference quotient, \(h\ne0\) and \(h\ne a-x\). These restrictions do not prevent \(h\) from approaching zero: because \(x\ne a\), sufficiently small increments leave \(x+h\) different from \(a\).
For example, if \(|h|<|x-a|/2\), then the triangle inequality gives \(|x+h-a|\ge |x-a|-|h|>|x-a|/2>0\). So all nonzero \(h\) in this small interval are valid increments. This makes the limiting calculation legitimate around every allowed input.
Deriving the Formula
At an input \(x\ne a\), substitute the function values into the difference quotient. The following calculation keeps the restrictions in view and simplifies the quotient before evaluating its limit.
For \(h\ne0\), with \(x+h\ne a\), use \(f(x+h)=c/(x+h-a)\) and \(f(x)=c/(x-a)\).
Use the common denominator \((x+h-a)(x-a)\) in the numerator, then divide by \(h\).
The numerator difference is \(-h\), so the nonzero factor \(h\) cancels.
As \(h\to0\), the remaining factor \(x+h-a\) approaches \(x-a\), which is nonzero.
The cancellation in the fourth line is valid because the difference quotient requires \(h\ne0\). We now take the limit of the simplified expression. Since \(x\ne a\), its denominator tends to the nonzero value \((x-a)^2\).
Why the Calculation Works
The definition involves a quotient by \(h\), which appears to become problematic as \(h\) tends to zero. The key is that the difference of the two function values contains a factor of \(h\). Combining the fractions reveals this factor explicitly, and canceling it leaves an expression that has a well-defined limit.
The remaining denominator has two factors: one is \(x-a\), and the other is \(x+h-a\). As \(h\) approaches zero, both approach \(x-a\). Since the chosen input is not \(a\), neither limiting factor is zero. This is why the simplified quotient approaches \(-c/(x-a)^2\) for every input in the domain.
The minus sign also has a concrete algebraic source: the numerator difference is \((x-a)-(x+h-a)=-h\). It is not an extra sign added by a differentiation convention. The difference quotient itself produces it.
Worked Examples
Worked Example: Differentiate \(6/(x-2)\) at \(x=5\)
Let \(f(x)=6/(x-2)\). Its excluded input is \(x=2\), so \(x=5\) is in its domain. For the difference quotient at \(5\), require \(h\ne0\) and \(5+h\ne2\), or \(h\ne-3\).
Taking the limit gives \(f'(5)=\lim_{h\to0}-2/(h+3)=-2/3\). Check using the derived formula: \(f'(5)=-6/(5-2)^2=-6/9=-2/3\). The difference quotient limit and the formula give the same value.
Worked Example: Differentiate \(-4/(x-1)\) at \(x=-1\)
Here \(f(x)=-4/(x-1)\), whose excluded input is \(1\). The input \(-1\) is allowed. In the difference quotient, \(h\ne0\) and \(-1+h\ne1\), so \(h\ne2\).
Therefore \(f'(-1)=\lim_{h\to0}-2/(h-2)=1\). Check with the general formula: \(f'(-1)=-(-4)/(-1-1)^2=4/4=1\). The sign is positive in this case because the coefficient \(c\) is negative.
Worked Example: Find the derivative function for \(3/(x+2)\)
Write the denominator as \(x-a\): here \(a=-2\) and \(c=3\). Thus the function is defined only for \(x\ne-2\). To see the formula directly, fix any such \(x\) and form the quotient. The increments must satisfy \(h\ne0\) and \(x+h\ne-2\).
Taking \(h\to0\) gives \(f'(x)=-3/(x+2)^2\) for \(x\ne-2\). As a numerical check, at \(x=0\) this gives \(f'(0)=-3/2^2=-3/4\). Substituting \(x=0\) into the simplified quotient gives \(-3/((h+2)2)\), whose limit is also \(-3/4\).
A Common Sign Error
A frequent mistake is to reverse the order of the two numerator terms while keeping the denominator unchanged. That changes the sign of the difference quotient. For instance, the correct numerator when combining fractions is \(c(x-a)-c(x+h-a)\), not \(c(x+h-a)-c(x-a)\).
If the incorrect numerator is used, the calculation would produce \(+c/(x-a)^2\). The correction is to preserve the order \(f(x+h)-f(x)\) from the definition. A useful check is to substitute a negative value of \(c\): the derived formula must still include the minus sign in front of \(c\), as the second worked example illustrates.
Keep the Excluded Input Visible
The formula \(f'(x)=-c/(x-a)^2\) is not a statement that the derivative exists at every real number. Both the original function and the derivative formula exclude \(x=a\). In particular, the difference quotient at \(a\) is not available because \(f(a)\) is undefined. Do not substitute \(x=a\) into the derivative expression or treat its denominator as though it were an ordinary nonzero number.
If \(c=0\), the same stated domain convention still applies: the expression \(0/(x-a)\) is defined for \(x\ne a\), not at \(a\). At every allowed input the formula gives \(f'(x)=0\). This does not assign a value to the original function at its excluded input.
The result is useful because it connects a simple rational expression to a derivative obtained from the defining limit, while making its domain restriction explicit. The essential habits are to identify excluded inputs first, use only valid increments in the difference quotient, simplify before taking the limit, and keep the sign and domain in the final formula.
Check Your Understanding
Answer each question using the difference-quotient reasoning and domain restrictions developed above.
- For \(f(x)=5/(x-3)\), which input is excluded, and what is \(f'(x)\) at an allowed input?
- Why must \(h\ne0\) and \(x+h\ne a\) when forming the quotient for \(c/(x-a)\)?
- What factor appears in the numerator after combining the two fractions, and why can it be canceled?
- For \(g(x)=-2/(x+1)\), find the excluded input and evaluate \(g'(1)\).
- What sign would result from reversing \(f(x+h)-f(x)\) in the numerator without changing anything else?
