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Differentiation methods

How to Use the Product Rule Without Losing a Term

A product derivative is not the product of derivatives. Each correct term changes one factor while preserving the other.

Direct answer

(fg)′ = f′g + fg′

Both factors may depend on x; a true constant factor can be pulled out.

When to use how to use the product rule

Choose the method from the top-level mathematical structure before manipulating symbols. Conditions are part of the task, not optional notes.

Ask first 1

Which structure controls the method?

What to do

Use the product rule when the top-level operation multiplies factors that depend on x.

Ask first 2

Which condition can change the answer?

What to do

Both factors may depend on x; a true constant factor can be pulled out.

Ask first 3

How should the result be checked?

What to do

Count one derivative term for each changing factor and ensure no term differentiates all factors at once.

Add and subtract the cross term

Start with the product difference quotient.

Add and subtract f(x+h)g(x), then split it into two quotients.

Continuity and the two derivative limits give f′g+fg′.

Apply the chain rule inside either factor when necessary.

1
(fg)′=f′g+fg′
2
[x²sin(x)]'=2x sin(x)+x²cos(x)
3
differentiate one factor per term

Worked examples

ddx[x2 sin⁡x]\frac{\mathrm{d}}{\mathrm{d}x}\left[x^{2}\,\sin x\right]
2x sin(x)+x²cos(x)

One term for each changing factor.

d/dx [x exp(x)]
ex (x+1)e^{x}\,\left(x + 1\right)

Factor exp(x) after applying the rule.

ddx[(x2+1) cos⁡(3 x)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^{2} + 1\right)\,\cos\left(3\,x\right)\right]
2x cos(3x)−3(x²+1)sin(3x)

The second factor also needs the chain rule.

Common mistakes

  • • Using f′g′.
  • • Writing only one term.
  • • Changing both factors in one term.
  • • Forgetting an inner chain rule.

Error clinic: locate the first invalid step

Incorrect attempt

[x²sin(x)]'=2x cos(x)

Why it fails

This multiplies factor derivatives and loses both product terms.

Correction

2x sin(x)+x²cos(x)

Incorrect attempt

[x exp(x)]'=exp(x)

Why it fails

Only x was differentiated.

Correction

exp(x)+x exp(x)

Practice the same idea with a changed structure

Name the rule and its conditions before revealing the reference answer. Explain every sign and factor.

d/dx [x exp(x)]

Check before solving

Factor exp(x) after applying the rule.

Reveal the reference answer

Reference answer

ex (x+1)e^{x}\,\left(x + 1\right)
ddx[(x2+1) cos⁡(3 x)]\frac{\mathrm{d}}{\mathrm{d}x}\left[\left(x^{2} + 1\right)\,\cos\left(3\,x\right)\right]

Check before solving

The second factor also needs the chain rule.

Reveal the reference answer

Reference answer

2x cos(3x)−3(x²+1)sin(3x)

Try the worked examples in the calculator

Each link opens the matching calculator with the expression, variable and required conditions already filled in.

Derivadas:x2 sin⁡xx^{2}\,\sin xDerivadas:x exx\,e^{x}
Open the calculator without a preset: Derivadas

Frequently asked questions

Why is f′g′ wrong?+

A small product change has two first-order contributions; the simultaneous change is second order.

What about three factors?+

Differentiate one factor at a time and preserve the other two, producing three terms.

Should I expand first?+

Only when expansion genuinely simplifies the derivative and preserves the domain.

Conclusion

Create one term per changing factor, preserve the other factors, and apply inner rules within each term.

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