Why a Composite Derivative Can Be Zero
A composite function changes through two linked processes. As \(x\) changes, the inner function \(g(x)\) changes; then the outer function \(F\) responds to that change. The chain rule records both rates:
This product can be zero for either of two reasons. The inner function may have derivative zero, so its output is not changing instantaneously. Or the outer function may have derivative zero at the current inner output, so the outer function is locally unresponsive there. Both factors can also be zero at once.
These are different sources of a zero derivative. To find every such point, check both factors rather than stopping as soon as one source appears.
The Zero-Product Test for Compositions
At any point where the chain rule applies, the derivative is a product of two real numbers. A product of two real numbers is zero exactly when at least one factor is zero. Thus we can find the zeros by solving the two factor equations separately.
Justification. By the chain rule, the derivative at \(a\) is \(F'(g(a))g'(a)\). If \(F'(g(a))=0\), this product is zero; if \(g'(a)=0\), it is also zero. Conversely, if the product is zero, the zero-product property of real numbers says that at least one factor is zero. This proves both directions of the equivalence.
The differentiability conditions matter: the chain-rule product gives the derivative only when \(g\) is differentiable at the input and \(F\) is differentiable at the corresponding inner output. The theorem does not claim that every composition has a derivative at every point.
A Reliable Method
For a composition \(F(g(x))\), first write its derivative in factored chain-rule form. Then solve \(g'(x)=0\) and \(F'(g(x))=0\) separately, keeping only values where the relevant functions and derivatives are defined. Finally, combine the two sets of solutions and check the derivative at each candidate.
Write the expression as \(F(g(x))\), naming the inner function \(g\) and the outer function \(F\).
Keep the factors \(F'(g(x))\) and \(g'(x)\) visible.
Solve \(g'(x)=0\), then solve \(F'(g(x))=0\).
List the distinct solutions, indicate their source, and verify them in the derivative.
Worked Example: A Polynomial Composition with Both Sources
Worked Example: A Polynomial Composition with Both Sources
Find all points where \(y=(x^2-1)^3\) has derivative zero, and classify the source of each zero.
Set \(g(x)=x^2-1\) and \(F(u)=u^3\). Then \(g'(x)=2x\) and \(F'(u)=3u^2\). The chain rule gives
The inner-source equation is \(g'(x)=2x=0\), so \(x=0\). The outer-source equation is \(F'(g(x))=3(x^2-1)^2=0\). A square is zero exactly when its base is zero, so \(x^2-1=0\), giving \(x=1\) or \(x=-1\).
At \(x=0\), \(g'(0)=0\), while \(F'(g(0))=F'(-1)=3\), so this is an inner-source zero only. At \(x=1\), \(g'(1)=2\) and \(F'(g(1))=F'(0)=0\). At \(x=-1\), \(g'(-1)=-2\) and \(F'(g(-1))=F'(0)=0\). Thus \(x=1\) and \(x=-1\) are outer-source zeros.
Check by substitution into the derivative:
The complete set is \(x=-1,0,1\), with the classifications above.
Worked Example: Zeros from a Trigonometric Outer Function
Worked Example: Zeros from a Trigonometric Outer Function
Find all real points where \(y=\sin((x-2)^2)\) has derivative zero, and identify the source of each zero.
Let \(g(x)=(x-2)^2\) and \(F(u)=\sin u\). Then \(g'(x)=2(x-2)\) and \(F'(u)=\cos u\). Therefore
The inner-source equation \(2(x-2)=0\) gives \(x=2\). For an outer-source zero, solve \(\cos((x-2)^2)=0\). Cosine is zero when its argument is \(\frac{\pi}{2}+k\pi\), where \(k\) is an integer. Since \((x-2)^2\geq 0\), the allowable values here are \(k=0,1,2,\ldots\). Hence
At these outer-source points the square root is positive, so \(x\ne2\) and the inner factor \(2(x-2)\) is nonzero. The points \(x=2\) and \(x=2\pm\sqrt{\frac{\pi}{2}+k\pi}\) therefore account for all zeros, with \(x=2\) from the inner derivative and the two families from the outer derivative.
For a check, at \(x=2\) the derivative has first factor \(2(2-2)=0\). At \(x=2\pm\sqrt{\frac{\pi}{2}+k\pi}\), the cosine factor is \(\cos(\frac{\pi}{2}+k\pi)=0\). Thus the derivative vanishes at every listed value.
Worked Example: When Only the Inner Derivative Can Vanish
Worked Example: When Only the Inner Derivative Can Vanish
Find the zero-derivative points of \(y=e^{x^2-4}\), and explain why there is no outer-source zero.
Take \(g(x)=x^2-4\) and \(F(u)=e^u\). Their derivatives are \(g'(x)=2x\) and \(F'(u)=e^u\). By the chain rule,
The inner-source equation \(2x=0\) gives \(x=0\). The outer-source equation would be \(e^{x^2-4}=0\), which has no real solutions because an exponential is positive for every real exponent. Therefore \(x=0\) is the only zero-derivative point, and it comes from the inner derivative.
Check: \(y'(0)=2(0)e^{-4}=0\). At every \(x\ne0\), both \(2x\) and \(e^{x^2-4}\) are nonzero, so their product is nonzero. This confirms there are no other zeros.
Common Mistake: Checking Only One Factor
A common error is to find where the inner derivative is zero and then assume those are all the zero-derivative points. In the polynomial example, that approach would solve \(2x=0\) and report only \(x=0\). But the full derivative is \(6x(x^2-1)^2\), and its outer factor also vanishes at \(x=1\) and \(x=-1\). The correction is to set each chain-rule factor equal to zero and combine the solutions.
The reverse mistake is also possible: solving only \(F'(g(x))=0\) misses zeros contributed by \(g'(x)\). Keep the chain-rule factors visible until both equations have been considered. Do not divide by a factor that might itself be zero; doing so could discard exactly the points being sought.
What a Zero Derivative Does—and Does Not—Tell You
A zero derivative says that the instantaneous rate of change of the composite is zero at that point. It does not, by itself, guarantee a local maximum or minimum. For instance, \(y=x^3\) has derivative \(3x^2\), which is zero at \(x=0\), even though the function increases on both sides of that point. The classification into inner-source and outer-source zeros explains how the chain-rule product becomes zero; it does not alone describe the surrounding behavior of the function.
It is also important not to confuse a zero of \(F'\) with a zero of \(F'(g(x))\) without checking the input. The outer derivative must be evaluated at the inner function’s value. In other words, the outer-source equation is \(F'(g(x))=0\), not simply an equation in which \(x\) is substituted directly into \(F'\).
Check Your Understanding
For each question, use the chain-rule factors to distinguish inner-source and outer-source zeros.
- For \(y=(x^2+1)^2\), find all zero-derivative points and classify their sources.
- For \(y=\cos(3x)\), write the two factor equations for possible derivative zeros. Which factor can vanish?
- For \(y=(x^2-4)^3\), find all zero-derivative points and identify any point where both factors vanish.
- Why must the outer-source equation be written using \(F'(g(x))\), rather than \(F'(x)\)?
- Does \(f'(a)=0\) guarantee that \(f\) has a local maximum or minimum at \(a\)? Give a reason.
