Derivatives · Definition

Product rule

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Before you start you need to know

The product rule differentiates \(fg\) using \(f\), \(g\) and their derivatives, without multiplying the derivatives together.

Definition

Rule giving the derivative of a product of two differentiable functions: \((fg)^{\prime}=f^{\prime}g+fg^{\prime}\). It is not the product of the derivatives.

\[(f\,g)'=f'\,g+f\,g'\]

The derivative of a product is not the product of the derivatives. The right rule is \((fg)'=f'g+fg'\): each factor contributes its change while the other stays still.

Intuition with a rectangle of sides \(f\) and \(g\): if both change a little, the area changes by two long strips (\(f'g\) and \(fg'\)) plus a tiny corner that vanishes in the limit.

A common mistake is differentiating each factor and multiplying; check with \(x\cdot x\): it gives \(2x\), not \(1\).

Think about it

Differentiate \(x^2\sin x\).

\(2x\sin x+x^2\cos x\).

Exercises 10

Differentiate the following functions

  1. \(\displaystyle f(x)=x^{2} \sin {x}\)
  2. \(\displaystyle f(x)=\left(3 x + 1\right) e^{x}\)
  3. \(\displaystyle f(x)=x \ln{\left(x \right)}\)
  4. \(\displaystyle f(x)=\left(x^{2} + 1\right) \cos {x}\)
  5. \(\displaystyle f(x)=e^{x} \sin {x}\)
  6. \(\displaystyle f(x)=x^{3} e^{x}\)
  7. \(\displaystyle f(x)=\left(x + 2\right) \cos {x}\)
  8. \(\displaystyle f(x)=\left(2 x - 1\right) e^{x}\)

Problems

  1. What is the mistake in this statement? "\((x^{2}\sin x)'=2x\cos x\)"
  2. Differentiate \(f(x)=x \sin {x}\, e^{x}\) by applying the product rule twice.

References

  • James Stewart, Calculus: Early Transcendentals, 8.ª edición, Cengage Learning, 2015.