Why a Linear Inner Function Matters
A power such as \(x^7\) can be differentiated directly with the power rule. An expression such as \((4x-1)^7\) has an extra layer: the seventh power is applied not to \(x\), but to the linear expression \(4x-1\). The chain rule accounts for how quickly that inner expression changes as \(x\) changes.
The extra factor is easy to overlook. It is also essential. Since \(4x-1\) changes four times as quickly as \(x\), differentiating the outer power alone does not give the correct derivative. The derivative must include the inner function’s derivative, \(4\).
The Rule for a Power of a Linear Expression
A linear function has the form \(u(x)=ax+b\), where \(a\) and \(b\) are constants. Its derivative is \(u'(x)=a\). The constant term \(b\) contributes no derivative; the coefficient \(a\) determines the rate of change.
For a positive integer \(n\), the power rule gives the derivative of the outer function \(F(u)=u^n\) as \(F'(u)=nu^{n-1}\). Applying the chain rule to the composition \(F(u(x))=(ax+b)^n\) gives the following formula.
The two factors \(n\) and \(a\) have different sources. The factor \(n\) comes from differentiating the outer power. The factor \(a\) comes from differentiating the inner linear function. Keep both factors, including the sign of \(a\).
Why the Inner Derivative Appears
The chain rule explains the method, but the appearance of the inner derivative can also be seen directly from the difference quotient. Let \(F\) be differentiable at \(u=ax+b\), and first suppose \(a\ne 0\). For an increment \(h\ne 0\), the change in the inner expression is \(ah\). Therefore, the difference quotient for the composite function can be rewritten as follows:
As \(h\) tends to zero, \(ah\) also tends to zero. By the definition of \(F'(u)\), the quotient \(\frac{F(u+ah)-F(u)}{ah}\) tends to \(F'(u)\). Thus the composite’s difference quotient tends to \(aF'(u)\). If \(a=0\), the inner expression is constant, so the composite is constant and has derivative zero. In either case, the derivative is \(F'(ax+b)a\), which is the chain rule for a linear inner function.
Now take \(F(u)=u^n\). Since \(n\) is a positive integer, the power rule gives \(F'(u)=nu^{n-1}\). Substituting \(u=ax+b\) and multiplying by \(a\) yields the stated formula. This derivation shows why the inner derivative is a factor: an increment \(h\) in \(x\) produces an increment \(ah\) in the input to the outer function.
A Reliable Calculation Process
For a power of a linear expression, identify the outer function and the inner function before differentiating. Then find each derivative and multiply them. This makes the roles of the exponent and the linear coefficient visible.
Write \(u=ax+b\), keeping the sign of every term.
For \(u^n\), the derivative with respect to \(u\) is \(nu^{n-1}\).
For \(u=ax+b\), this derivative is \(a\).
Multiply the outer derivative by \(a\), then write the result as a function of \(x\).
Worked Examples
Worked Example: A Seventh Power with a Positive Coefficient
Differentiate \(f(x)=(4x-1)^7\).
The inner function is \(u(x)=4x-1\), so \(u'(x)=4\). The outer function is \(F(u)=u^7\), so \(F'(u)=7u^6\). Multiply the outer derivative by the inner derivative:
Check the factors: lowering the exponent gives \(7\), and differentiating \(4x-1\) gives \(4\), so the coefficient is \(7\cdot4=28\). At \(x=1\), the result gives \(f'(1)=28(4-1)^6=28\cdot729=20412\). The inner derivative is constant and positive, consistent with the positive factor in the result.
Worked Example: A Negative Inner Coefficient
Differentiate \(g(x)=(3-2x)^5\). Here the inner function is \(u(x)=3-2x\), whose derivative is \(u'(x)=-2\). The outer derivative is \(5u^4\). Therefore,
Check the sign and coefficient: the outer derivative contributes \(5\), while the inner derivative contributes \(-2\), and \(5\cdot(-2)=-10\). At \(x=0\), \(g'(0)=-10\cdot3^4=-10\cdot81=-810\). The negative sign is required because the inner expression decreases as \(x\) increases.
Worked Example: Checking by Expanding a Square
Differentiate \(h(x)=(5x+2)^2\), and check the result by expanding first.
Using the chain rule, the outer derivative is \(2(5x+2)\), and the inner derivative is \(5\). Thus
For an independent check, expand the square and differentiate the resulting polynomial:
The expanded calculation matches the chain-rule result. The check also confirms that the factor \(5\) from the inner expression is necessary.
Worked Example: Checking a Fourth Power by Expansion
Differentiate \(p(x)=(1-3x)^4\). The inner derivative is \(-3\), and the outer derivative is \(4u^3\). Consequently,
To check, expand the original expression and differentiate each term:
Expanding the chain-rule result gives the same polynomial:
The two calculations agree term by term.
Common Mistake: Omitting the Inner Derivative
A frequent error is to lower the exponent and multiply only by the old exponent. For example, the incorrect calculation
has treated \(4x-1\) as though its derivative were \(1\). But \(\frac{d}{dx}(4x-1)=4\), so the corrected calculation is
Another error is to lose a negative sign. For \((3-2x)^5\), the inner derivative is \(-2\), not \(2\), so the derivative is \(-10(3-2x)^4\). A useful check is to differentiate the entire inner expression, including its coefficient and sign, before multiplying.
What the Formula Tells You
The expression \((ax+b)^n\) is defined for every real \(x\) when \(n\) is a positive integer, so there is no restriction on \(x\) for this formula. If \(a=0\), the expression becomes the constant \(b^n\), and the formula correctly gives zero. If \(n=1\), it gives \(\frac{d}{dx}(ax+b)=a\), the derivative of the linear function itself.
Most importantly, the coefficient \(a\) is not an optional adjustment. The derivative describes the rate of change with respect to \(x\), and the inner expression changes at rate \(a\). A negative \(a\) reverses the sign of the derivative factor; a larger \(|a|\) scales its magnitude. This is why checking the inner derivative is part of the calculation, not merely a final detail.
Check Your Understanding
For each question, identify the inner derivative before simplifying the derivative.
- Differentiate \((6x+1)^4\).
- Differentiate \((2-5x)^3\), and explain where the negative sign comes from.
- Differentiate \((7x-4)^2\), then check by expanding the square.
- What is the derivative of \((9)^5\) when it is viewed as a constant function of \(x\)?
- In differentiating \((ax+b)^n\), which factor comes from the outer power, and which comes from the inner function?
