Why Check Derivatives From First Principles?
A derivative formula can make calculations efficient, but it is useful to know what the formula says at its foundation. The derivative at an input is the limit of nearby average rates of change. For a polynomial, forming that limit directly shows how the familiar constant, linear, and quadratic derivative formulas arise.
This tutorial uses the difference quotient at a general input. The goal is not to replace the polynomial differentiation rules, but to verify these basic cases from the definition and understand the algebra that makes the limit work.
The Derivative as a Limit
Let \(f\) be a function defined at \(x\) and at inputs sufficiently close to \(x\). For a nonzero increment \(h\), the difference quotient is the change in output divided by the change in input:
It is the average rate of change from \(x\) to \(x+h\). If this quotient approaches a finite value as \(h\) approaches zero through nonzero values, that value is the derivative at \(x\).
The condition \(h\ne 0\) matters because the denominator is \(h\). The expression is not evaluated by setting \(h=0\) in the quotient. Instead, we simplify the quotient where it is defined and then find its limit.
Every polynomial is defined for every real input, so for any fixed \(x\), the values \(f(x+h)\) and \(f(x)\) are available for all real \(h\). We will use ordinary algebra to simplify their difference, cancel a factor of \(h\) only while \(h\ne 0\), and then take the limit.
Worked Example: A Constant Polynomial
Worked Example: A Constant Polynomial
Let \(f(x)=7\). Verify its derivative at an arbitrary real input \(x\) from the definition.
For every nonzero \(h\), the value of the function at \(x+h\) is also \(7\). Therefore, the difference quotient is
Thus, as \(h\) approaches zero through nonzero values, the quotient remains \(0\). It follows that
This holds for every real \(x\). As a check, the simplified difference quotient is exactly \(0\) for every permitted \(h\), so its limit is exactly the derivative claimed.
The reason a constant has zero derivative is visible in the definition: changing the input does not change the output. The numerator is therefore zero, and the average rate of change is zero for every nonzero increment.
Worked Example: A Linear Polynomial
Worked Example: A Linear Polynomial
Let \(f(x)=4x-3\). Find the derivative at a general real input \(x\) using the difference quotient.
First, evaluate the function at the shifted input:
Now subtract the original function value and divide by \(h\), with \(h\ne 0\):
The quotient is \(4\) for every nonzero \(h\). Hence
The result does not depend on \(x\), which matches the constant rate of change of a line. To check the result directly, the simplified quotient is \(4\) throughout its domain \(h\ne0\), so its limit is indeed \(4\).
The same calculation works for any linear polynomial. Its coefficient of \(x\) determines the change in output for each unit change in input; its constant term cancels when the two function values are subtracted.
Worked Example: A Quadratic Polynomial
Worked Example: A Quadratic Polynomial
Let \(f(x)=x^2\). Verify the derivative at an arbitrary real input \(x\).
Substitute \(x+h\) into the function and expand the square:
Form the difference quotient. The denominator requires \(h\ne0\):
Because \(h\ne0\), cancelling the common factor \(h\) is valid in the difference quotient. The simplified expression has a limit as \(h\) approaches zero:
As a check, the quotient equals \(2x+h\) for every nonzero \(h\), and this expression approaches \(2x\). Thus the derivative at every real input is \(2x\), as claimed.
The General Quadratic Calculation
The same steps verify the derivative of every quadratic polynomial, including a possible linear term and constant term. Let \(f(x)=ax^2+bx+c\), where \(a\), \(b\), and \(c\) are fixed real numbers. For any real \(x\) and any nonzero \(h\), first expand the shifted input:
Subtract \(f(x)=ax^2+bx+c\). The terms that do not involve \(h\) cancel:
Now divide by \(h\), which is permitted because \(h\ne0\), and factor the result:
As \(h\) approaches zero, the term \(ah\) approaches zero, while \(2ax+b\) is fixed with respect to \(h\). Therefore, the limit exists and equals \(2ax+b\).
Why the Calculation Works
The key is that the change in a polynomial can be expressed with a factor of \(h\). In the quadratic calculation, the terms that remain after subtraction are \(2axh+ah^2+bh\). Each contains \(h\), so the common factor can be cancelled against the denominator before taking the limit.
After cancellation, the quotient is \(2ax+b+ah\). The only part that still depends on the increment is \(ah\), and that part tends to zero. This explains why the derivative is \(2ax+b\), rather than a quotient that is undefined at \(h=0\). The limit is taken after the algebraic simplification.
The general calculation also contains the simpler cases. If \(a=0\), it gives the linear polynomial \(bx+c\), with derivative \(b\). If \(a=0\) and \(b=0\), it gives the constant polynomial \(c\), with derivative \(0\). These conclusions agree with the direct difference-quotient calculations.
A Common Mistake: Substituting Zero Too Soon
For \(f(x)=x^2\), a mistaken calculation might set \(h=0\) immediately in the original difference quotient:
This does not show that the derivative is zero or that it fails to exist. The quotient is undefined at \(h=0\), and \(0/0\) has no numerical value. The derivative asks for the limit of the quotient at nonzero \(h\), not its value at \(h=0\).
The correction is to simplify first, while \(h\ne0\):
Now take the limit of the simplified expression: \(\lim_{h\to0}(2x+h)=2x\). Cancelling \(h\) does not mean claiming that \(h\ne0\) at the limit. It means that the original quotient and the simplified expression agree for every nonzero \(h\), which is exactly the range of values used to determine the limit.
Why First-Principles Checks Matter
For constant, linear, and quadratic polynomials, the difference quotient makes the derivative formula concrete: subtracting nearby outputs removes the unchanged terms, and the remaining increment factor can be cancelled. This is useful both for understanding the meaning of the derivative and for checking that a formula describes the limiting rate of change at a general input.
When using the definition, keep the order of operations clear: form the quotient for \(h\ne0\), simplify it, and only then take the limit as \(h\) approaches zero. This order avoids the undefined expression \(0/0\) and shows precisely where the derivative value comes from.
Check Your Understanding
For each question, use the difference quotient definition and keep \(h\ne0\) until after simplifying.
- For \(f(x)=9\), what is the difference quotient at a general input \(x\), and what is its limit as \(h\to0\)?
- For \(f(x)=-3x+5\), expand \(f(x+h)\), simplify the quotient, and state the derivative at a general input.
- For \(f(x)=2x^2\), simplify the difference quotient and find its limit.
- For \(f(x)=x^2+6x-1\), show which terms cancel after subtracting \(f(x)\), then determine the derivative.
- Why is substituting \(h=0\) into the original difference quotient not a valid way to find the derivative?
