Why These Expressions Need Two Derivatives
A powered polynomial such as \((x^3-2x+1)^6\) has an outer operation and an inner expression. The outer operation raises a quantity to the sixth power; the inner quantity is the polynomial \(x^3-2x+1\). Differentiating correctly requires accounting for both layers.
The power rule tells us how to differentiate a power of a variable. The chain rule explains what changes when the quantity being raised to a power is itself a function of \(x\). Together, they give a compact method for differentiating powers of polynomial expressions without expanding them first.
The Rule for a Positive Integer Power
Let \(p(x)\) be a polynomial and let \(n\) be a positive integer. The expression \([p(x)]^n\) is a composite function: its inner function is \(p(x)\), and its outer function raises an input to the \(n\)th power. Since polynomials are differentiable for every real \(x\), the chain rule applies at every real input.
The factor \(n[p(x)]^{n-1}\) comes from differentiating the outer power while treating its input as a single quantity. The remaining factor, \(p'(x)\), comes from differentiating that input with respect to \(x\). Both factors are essential.
For example, in \((x^3-2x+1)^6\), the inner polynomial is \(p(x)=x^3-2x+1\), so \(p'(x)=3x^2-2\). The exponent decreases from 6 to 5, and the derivative of the inner polynomial multiplies the result.
Why the Rule Works
The justification is a direct application of the chain rule and the power rule. Define an outer function \(g(u)=u^n\), where \(u\) is its input, and let the inner function be \(p(x)\). Then \([p(x)]^n=g(p(x))\).
For a positive integer \(n\), the power rule gives \(g'(u)=nu^{n-1}\). The chain rule says that the derivative of \(g(p(x))\) is \(g'(p(x))p'(x)\). Substituting \(p(x)\) into the outer derivative gives the stated formula:
The conditions are met: \(p\) is a polynomial, so it is differentiable for every real \(x\), and \(g(u)=u^n\) is differentiable for every real \(u\) when \(n\) is a positive integer. Thus the formula holds for every real \(x\), including points where \(p(x)=0\).
This derivation also explains the role of each factor. The exponent decreases because of the outer power rule; the derivative \(p'(x)\) appears because the input to that outer function changes as \(x\) changes.
A Reliable Three-Step Method
Write down the full expression being raised to a power, and determine its derivative.
Bring the exponent down as a coefficient and reduce the exponent by one, keeping the inner polynomial in parentheses.
Attach the derivative of the polynomial as a factor. Then simplify only if doing so preserves the structure clearly.
Keeping the polynomial grouped in parentheses helps distinguish the inner function from the outer power. It also makes the required inner derivative factor easier to see.
Worked Examples
Worked Example: A Sixth Power of a Cubic
Differentiate \(y=(x^3-2x+1)^6\).
First identify the inner polynomial and differentiate it term by term:
Apply the rule with \(n=6\). The exponent becomes \(6-1=5\), and the inner derivative is included as a factor:
Check: At \(x=0\), the inner polynomial has value \(p(0)=1\), and its derivative has value \(p'(0)=-2\). The formula gives \(6(1)^5(-2)=-12\). Independently, view \(p(x)^6\) as a product of six copies of \(p(x)\). At \(x=0\), the product rule gives six terms, each with one factor \(p'(0)=-2\) and five factors \(p(0)=1\), so the derivative is \(6(-2)(1)^5=-12\), as required.
Worked Example: A Quadratic Raised to the Third Power
Differentiate \(y=(3x^2-5)^3\).
The inner polynomial and its derivative are
Bring down the outer exponent, reduce it by one, and multiply by \(6x\):
Check: At \(x=1\), \(p(1)=3-5=-2\) and \(p'(1)=6\). The derivative formula gives \(3(-2)^2(6)=3(4)(6)=72\). To check by the product rule, write \(p^3=p\cdot p\cdot p\). Its derivative is \(p'p^2+pp'p+ p^2p'=3p'p^2\). At \(x=1\), this is \(3(6)(4)=72\), matching the result.
Worked Example: Checking by Expanding a Square
Differentiate \(y=(x^2+1)^2\). Here the inner polynomial is \(p(x)=x^2+1\), so \(p'(x)=2x\). Applying the rule gives
For an independent check, expand the original expression and differentiate the resulting polynomial:
Factoring the expanded derivative gives \(4x^3+4x=4x(x^2+1)\), exactly the same answer. Expanding is not necessary for the method; here it serves as a check.
When the Inner Polynomial Is Zero
The formula remains valid when \(p(x)=0\). For a positive integer exponent \(n\), the expression \([p(x)]^n\) is still a polynomial expression, and its derivative is defined at such inputs.
If \(n>1\) and \(p(a)=0\), the formula gives \(n[p(a)]^{n-1}p'(a)=0\), because the factor \([p(a)]^{n-1}\) is zero. For instance, at \(x=1\), \(x^3-2x+1=0\), so the derivative of \((x^3-2x+1)^6\) is zero there. The rule does not require dividing by \(p(x)\), so there is no problem at a zero of the inner polynomial.
The exponent \(n=1\) is also covered: the formula becomes \(1[p(x)]^0p'(x)=p'(x)\), using \(p(x)^0=1\). A zeroth power is a separate constant case: \([p(x)]^0=1\) wherever that expression is defined, and its derivative is zero. The main rule above is stated for positive integer exponents.
A Common Mistake: Forgetting the Inner Derivative
A frequent error is to apply the outer power rule and stop before multiplying by the derivative of the polynomial. For example, an incorrect attempt at differentiating \((3x^2-5)^3\) might be
This calculation accounts for the exponent but treats the inner expression as though it changed at a rate of \(1\). In fact, the derivative of \(3x^2-5\) is \(6x\), so that rate must be included:
A useful check is to compare the proposed answer with the three-step method: did the outer power produce the coefficient and reduced exponent, and did the inner polynomial contribute its derivative as a factor? If the latter factor is absent, the chain rule has not been completed.
Why This Method Is Useful
Expanding a powered polynomial before differentiating can create many terms. For example, a sixth power of a cubic would expand into a polynomial with several powers of \(x\). The chain-rule form is usually shorter and keeps the structure visible: one factor records the outer power, and another records how the inner polynomial changes.
The method also makes it easier to check each part of the derivative. You can verify the exponent and coefficient from the outer power rule, then verify the inner factor by differentiating the polynomial. This is especially valuable when the polynomial has several terms or the outer exponent is large.
Check Your Understanding
For each question, identify the inner polynomial before applying the rule.
- Differentiate \((x^2-4x+3)^5\).
- Differentiate \((2x^3+x)^4\), showing the derivative of the inner polynomial.
- For \(y=(x^2-1)^3\), find \(y'\) and evaluate it at \(x=1\).
- What factor is missing from the incorrect derivative \(4(x^2+3)^3\) for \((x^2+3)^4\)?
- Explain why the power rule for a positive integer exponent still applies at an input where the inner polynomial equals zero.
