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Chain-rule mistakes
Why Do You Multiply by the Inner Derivative?
The inner derivative is not an extra decoration. It converts the outer rate, measured per unit of the inner input, into a rate per unit of x.

Direct answer
The rule applies when one differentiable function is evaluated inside another; every nontrivial nested layer contributes a factor.
Rate argument: connect the two changes
Let u=g(x) and y=f(u). The outer derivative dy/du measures how y changes when u changes.
The inner derivative du/dx measures how u changes when x changes. Multiplying (dy/du)(du/dx) converts the outer rate into dy/dx.
For y=sin(x^2), cosine measures the change per unit of x^2, while 2x measures how quickly x^2 changes. The result is 2x cos(x^2).
A useful audit is to count nontrivial layers before simplifying. If a layer changes with x but contributes no derivative factor, the solution is incomplete.
Worked examples
cos(x^2) is the outer derivative and 2x is the inner derivative.
The exponential remains, while the linear inside contributes 3.
Differentiate the cube, preserve x^2+1, then multiply by 2x.
Common mistakes
- • Stopping at cos(x^2) after differentiating sin(x^2).
- • Replacing the inner expression instead of preserving it inside the outer derivative.
- • Multiplying by the inner expression rather than its derivative.
- • Simplifying early and losing track of which layer supplied each factor.
Try the worked examples in the calculator
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Open the calculator without a preset: 链式法则计算器References
Frequently asked questions
Do I multiply by the inner function or its derivative?+
Multiply by the derivative of the inner function. The unchanged inner function stays inside the outer derivative.
What if the inner derivative equals one?+
The factor is still conceptually present, but multiplying by one does not change the written result.
Does every nested layer add a factor?+
Every nontrivial differentiable layer adds a derivative factor. Constant or identity layers may simplify to zero or one.
Conclusion
Multiply by the inner derivative because the outer input does not usually change one-for-one with x. Keep the inside visible, differentiate each layer, and multiply the resulting rates.
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