From a Difference Quotient to a Derivative
A difference quotient compares the change in a function’s output with the change in its input. To find a derivative from its definition, the key calculation is to simplify this quotient before taking its limit. For linear and polynomial functions, expansion and factoring often reveal a factor of the increment. For a reciprocal function, combining fractions exposes that factor. For a radical function, multiplying by a conjugate can do the same job.
The increment must not be set equal to zero inside the quotient: the denominator would then be zero. Instead, simplify the quotient for nonzero increments near zero, and then evaluate the limit of the simplified expression.
A Reliable Calculation Pattern
Most calculations in this workshop follow the same sequence. First substitute \(a+h\) into the function and calculate \(f(a)\). Then form the numerator \(f(a+h)-f(a)\). Simplify that numerator until the vanishing factor \(h\) is visible, cancel it only when \(h\ne 0\), and take the limit of what remains.
Write \(f(a+h)\) and \(f(a)\) accurately before combining them.
Expand, combine fractions, or use a conjugate, according to the function.
Once the numerator and denominator share a factor \(h\), cancel it for \(h\ne0\).
Evaluate the limit of the simplified expression as \(h\) approaches zero.
Worked Example: A Linear Function
Worked Example: A Linear Function
Find the derivative of \(f(x)=3x-5\) at \(a=2\) from the limit definition.
First calculate the function value at the point:
Now substitute \(2+h\) into the function:
Form and simplify the quotient, keeping \(h\ne0\):
Therefore, \(\displaystyle f'(2)=\lim_{h\to0}3=3\). The simplified quotient is constant, so its limit is 3. As a check, the original numerator is \(3h\), and dividing the output change \(3h\) by the input change \(h\) gives the constant rate 3 for every nonzero \(h\).
Why Cancellation Works in a Limit
In the definition, the difference quotient is not evaluated at \(h=0\). It is evaluated for nonzero \(h\) and then considered as \(h\) approaches zero. If algebra shows that the quotient equals a simpler expression for every allowed \(h\ne0\), those two expressions have the same limit as \(h\to0\). The cancellation does not assign a value to the original quotient at zero; it gives an equivalent expression on the punctured interval around zero.
For example, for \(f(x)=x^2\) at a number \(a\), the numerator factors as follows:
The factor \(h\) was canceled only under the condition \(h\ne0\). The expression that remains has limit \(2a\) as \(h\to0\). This illustrates why the method works: algebra replaces an expression undefined at \(h=0\) with an equal expression for nearby nonzero \(h\), whose limit can be read directly.
Worked Example: A Cubic Polynomial
Worked Example: A Cubic Polynomial
Find the derivative of \(f(x)=x^3\) at \(a=2\) from the limit definition.
Evaluate the function at \(2\), then at \(2+h\):
Subtract the function value and divide by \(h\):
Now take the limit:
Check the simplification by multiplying \(12+6h+h^2\) by \(h\): the result is \(12h+6h^2+h^3\), the numerator before division. Thus the quotient has been simplified correctly, and its limit is 12.
Worked Example: A Reciprocal Function
Worked Example: A Reciprocal Function
Find the derivative of \(f(x)=\frac{1}{x}\) at \(a=2\). The function is defined near \(2\), provided its input is not zero.
Calculate the two function values:
For the quotient, require \(h\ne0\) and \(2+h\ne0\), so \(h\ne-2\) as well. Combine the fractions in the numerator before dividing by \(h\):
As \(h\to0\), the denominator \(2(2+h)\) approaches \(4\), so
To check the sign, the numerator of the combined fractions is \(2-(2+h)=-h\), not \(h\). The negative sign therefore remains after canceling the common factor \(h\).
Worked Example: A Radical Function
Worked Example: A Radical Function
Find the derivative of \(f(x)=\sqrt{x}\) at \(a=4\). For real inputs, the function is defined when \(x\ge0\); the increments near \(h=0\) must satisfy \(4+h\ge0\).
Evaluate the function values and form the quotient:
The numerator is a difference of radicals. Multiply the numerator and denominator by the conjugate \(\sqrt{4+h}+2\). This does not change the quotient when \(h\ne0\): it multiplies it by 1.
The denominator in the final expression approaches \(2+2=4\), so
Check the conjugate step using \((u-v)(u+v)=u^2-v^2\): with \(u=\sqrt{4+h}\) and \(v=2\), the numerator becomes \((4+h)-4=h\), exactly the factor canceled from the denominator.
Common Pitfall: Canceling Terms Instead of Factors
Cancellation applies to common factors in a product, not to pieces of a sum. For instance, after expanding the cubic example, a student might incorrectly cancel the \(h\) in \(12h+6h^2+h^3\) against the denominator and write \(12+6h^2+h^3\). This is wrong: the numerator is a sum, so the common factor must first be factored from every term.
A quick check catches the error: multiplying the incorrect expression \(12+6h^2+h^3\) by \(h\) does not recover \(12h+6h^2+h^3\). Multiplying the correct expression \(12+6h+h^2\) by \(h\) does recover it.
Choosing the Algebraic Step
The function’s form suggests which operation is most useful. Expansion reveals powers of \(h\) in a polynomial difference. Combining fractions is essential for a reciprocal difference, because the numerator must be put over a common denominator before its common factor can be seen. For a radical difference, the conjugate turns the numerator into a difference of squares.
| Function form | Useful step | What to check |
|---|---|---|
| Linear or polynomial | Substitute and expand; factor \(h\) from the numerator. | Keep all terms when expanding. |
| Reciprocal | Combine the fractions in the numerator. | Track the order of subtraction and note excluded inputs. |
| Radical | Multiply by the conjugate. | Use the difference-of-squares identity correctly. |
Across all three forms, the central discipline is the same: preserve the difference quotient for nonzero increments, simplify it by valid algebra, and only then take the limit. The algebraic expression may change, but the limit calculation remains grounded in the original definition.
Check Your Understanding
For each problem, write the difference quotient, simplify it for nonzero \(h\), and evaluate its limit.
- Find the derivative of \(f(x)=5x+1\) at \(a=3\).
- Find the derivative of \(f(x)=x^2+3x\) at \(a=-1\).
- Find the derivative of \(f(x)=\frac1x\) at \(a=1\). State any restriction on \(h\) used in the quotient.
- Find the derivative of \(f(x)=\sqrt{x}\) at \(a=9\).
- Explain why canceling a common factor \(h\) in a difference quotient does not mean that the original quotient is defined at \(h=0\).
