Introduction
A parameter can make a familiar derivative rule feel less familiar. In an expression such as \((x+a)(x^2-a)\), both \(x\) and \(a\) appear, but if the derivative is with respect to \(x\), only \(x\) varies. The parameter \(a\) is held fixed during the differentiation.
This distinction matters when differentiating and when evaluating the result. First find a formula for the derivative in terms of \(x\) and the parameter. Then substitute the specified values, checking any restrictions on the original function.
Parameters and the Variable of Differentiation
A parameter is a fixed quantity that may vary between different instances of a problem. For example, when \(a=3\), the function \((x+a)(x^2-a)\) is one particular function of \(x\). When \(a=5\), it is a different function of \(x\). In either instance, \(a\) remains constant as \(x\) changes.
Thus, with respect to \(x\), \(\frac{d}{dx}(x+a)=1\), \(\frac{d}{dx}(x^2-a)=2x\), and \(\frac{d}{dx}a=0\). These facts follow because a constant has derivative zero and the derivative is taken with respect to \(x\).
The product rule states that if \(u\) and \(v\) are differentiable functions of \(x\), then \((uv)'=u'v+uv'\). The quotient rule states that if \(u\) and \(v\) are differentiable and \(v(x)\ne 0\), then \(\left(\frac{u}{v}\right)'=\frac{u'v-uv'}{v^2}\). The denominator condition is essential: the original quotient is not defined where \(v=0\).
Worked Example: Differentiating a Parameterized Product
Worked Example: Differentiating a Parameterized Product
Let \(f(x)=(x+a)(x^2-a)\), where \(a\) is fixed. Find \(f'(x)\), then evaluate \(f'(2)\) when \(a=3\).
Apply the product rule to the two factors. Their derivatives with respect to \(x\) are \(1\) and \(2x\), respectively:
Expand and collect terms:
Now substitute \(x=2\) and \(a=3\):
Check by substituting \(a=3\) into the original function first. Then \(f(x)=(x+3)(x^2-3)\), so the product rule gives \(f'(2)=(2^2-3)+(2+3)(2\cdot2)=1+20=21\). The two calculations agree.
Why the Parameter Stays Fixed
The rule that \(a\) is constant is not merely a convention for simplifying notation. It describes which quantity is changing in the derivative. A short difference-quotient calculation makes this explicit for the product above.
For a fixed real number \(a\), the derivative at \(x\), if it exists, is the limit of the change in the function divided by the change in \(x\). Let \(h\ne0\). The difference in the function values can be expanded as follows:
Divide by \(h\), which is nonzero, and take the limit as \(h\) tends to zero:
In this limit, \(x\) and \(a\) are fixed while \(h\) approaches zero. In particular, terms containing \(h\) vanish, but \(a\) does not. The result agrees with the product-rule calculation.
Worked Example: Differentiating a Parameterized Quotient
Worked Example: Differentiating a Parameterized Quotient
Let \(g(x)=\frac{x+a}{x-a}\), where \(a\) is fixed. Find \(g'(x)\), then evaluate \(g'(3)\) when \(a=1\).
The denominator \(x-a\) must be nonzero, so the function is defined only when \(x\ne a\). On that domain, the quotient rule applies with numerator \(u=x+a\) and denominator \(v=x-a\). Their derivatives are \(u'=1\) and \(v'=1\):
For \(a=1\), the input \(x=3\) is in the domain because \(3\ne1\). Substitute these values into the derivative:
Check by first setting \(a=1\): \(g(x)=\frac{x+1}{x-1}\). The quotient rule at \(x=3\) gives \(\frac{1(2)-4(1)}{2^2}=\frac{2-4}{4}=-\frac12\), as required.
Evaluating Derivatives and Checking the Domain
When a question gives both an input and a parameter, the order of operations helps keep the calculation reliable. First derive a formula with \(x\) and the parameter still present. Next check that the requested input is allowed by the original function. Only then substitute the values.
For a product of polynomial factors, there is no denominator restriction, so every real input is allowed. For a quotient, the denominator must be nonzero. In the example \(g(x)=\frac{x+a}{x-a}\), the restriction is \(x\ne a\). It applies to the original function, even if a later algebraic simplification seems to remove a factor.
Record that the derivative is with respect to \(x\), so each parameter is constant.
Use the product rule or quotient rule and retain the parameter in the formula.
For a quotient, verify that its original denominator is nonzero at the requested input.
Insert the given input and parameter values, then check the arithmetic.
Worked Example: Finding Parameter Values from a Derivative Condition
Worked Example: Finding Parameter Values from a Derivative Condition
For \(g(x)=\frac{x+a}{x-a}\), find all real values of \(a\) for which \(g'(2)=-2\).
The derivative formula is \(g'(x)=\frac{-2a}{(x-a)^2}\), and it applies only when \(x\ne a\). At \(x=2\), this requires \(a\ne2\). Use the derivative condition:
Since \(a\ne2\), the squared denominator is positive, so multiplying both sides by \((2-a)^2\) is valid:
Divide both sides by \(-2\), then expand and collect terms:
Factor the quadratic and solve:
Both values satisfy the domain condition \(a\ne2\). Check them in the derivative formula at \(x=2\): for \(a=1\), \(g'(2)=-2(1)/(2-1)^2=-2\); for \(a=4\), \(g'(2)=-2(4)/(2-4)^2=-8/4=-2\). Both are valid solutions.
A Common Sign and Domain Pitfall
A frequent error in the quotient rule is to reverse the subtraction in its numerator. For \(g(x)=\frac{x+a}{x-a}\), an incorrect calculation might be
This uses the numerator times the denominator’s derivative first, rather than the numerator’s derivative times the denominator. The quotient rule requires \(u'v-uv'\), in that order. The corrected calculation is
The sign can also be checked with a concrete choice: when \(a=1\) and \(x=3\), the correct derivative is \(-\frac12\), not \(+\frac12\).
There is also a domain pitfall when \(a=0\). In that case, the quotient is \(\frac{x}{x}=1\) for \(x\ne0\), and its derivative is zero for \(x\ne0\). But the original quotient is still undefined at \(x=0\). Simplifying the expression does not restore a value to the original function at a point where its denominator was zero. The derivative formula \(\frac{-2a}{(x-a)^2}\) likewise cannot be evaluated at \(x=a\).
Why These Calculations Matter
A derivative involving a parameter describes a family of functions rather than just one function. Its formula shows how the slope depends on both the input and the parameter. This makes it possible to compare members of the family, evaluate slopes at specified points, or find parameter values that meet a condition.
The reliable method is the same in each setting: specify the variable of differentiation, keep the parameters fixed during differentiation, preserve domain restrictions, and substitute only after the derivative formula is established. For a product, check that both product-rule terms are present. For a quotient, preserve the order \(u'v-uv'\) and verify that the denominator is nonzero where the derivative is evaluated.
Check Your Understanding
Answer each question, keeping the parameter fixed when differentiating with respect to \(x\).
- Find the derivative of \(p(x)=(x+b)(x^2+b)\) with respect to \(x\).
- For \(p(x)=(x+b)(x^2+b)\), evaluate \(p'(1)\) when \(b=2\).
- Find the derivative of \(q(x)=\frac{x+c}{x-c}\), and state its domain.
- Evaluate \(q'(4)\) when \(c=2\).
- For \(q(x)=\frac{x+c}{x-c}\), what restriction on \(c\) is needed to evaluate \(q'(3)\)?
