CalculusLesson 1 of 1 in the course Derivatives

Product Rule: Proof, Higher Derivatives, and Extensions

5 flashcards · 5 quiz questions · 5 match pairs · 4 fill-in sentences · English

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Summary

The product rule tells us how to differentiate a product of two differentiable functions. It matters whenever both factors depend on the variable: differentiating each factor separately and multiplying the results is not enough.

If uu and vv are differentiable functions of xx, then

ddx(uv)=dudxv+udvdx.\frac{d}{dx}(uv)=\frac{du}{dx}v+u\frac{dv}{dx}.

The rule says to differentiate one factor at a time, then add the results: first differentiate uu and keep vv unchanged; then keep uu unchanged and differentiate vv. Both terms are necessary.

All flashcards

Product rule
For differentiable functions uu and vv, (uv)′=u′v+uv′(uv)'=u'v+uv'.
Product-rule proof by limits
Add and subtract f(x)g(x+Δx)f(x)g(x+\Delta x) to split the product's difference quotient into two parts.
Second derivative of a product
For twice-differentiable uu and vv, (uv)′′=u′′v+2u′v′+uv′′(uv)''=u''v+2u'v'+uv''.
General Leibniz rule
The nnth derivative of uvuv is ∑k=0n(nk)u(n−k)v(k)\sum_{k=0}^{n}\binom{n}{k}u^{(n-k)}v^{(k)}.
Graded product rule for differential forms
For a kk-form α\alpha, d(α∧β)=dα∧β+(−1)kα∧dβd(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^k\alpha\wedge d\beta.

Quiz questions

  1. If uu and vv are differentiable, which formula gives the derivative of their product?Answer: (uv)′=u′v+uv′(uv)'=u'v+uv'
  2. In the limit proof for h(x)=f(x)g(x)h(x)=f(x)g(x), which expression is added and subtracted in the numerator to split the difference quotient?Answer: f(x)g(x+Δx)f(x)g(x+\Delta x)
  3. For twice-differentiable functions uu and vv, what is (uv)′′(uv)''?Answer: u′′v+2u′v′+uv′′u''v+2u'v'+uv''
  4. If differentiable vector functions f\mathbf f and g\mathbf g have a dot product, how is its derivative expressed?Answer: (f⋅g)′=f′⋅g+f⋅g′(\mathbf f\cdot\mathbf g)'=\mathbf f'\cdot\mathbf g+\mathbf f\cdot\mathbf g'
  5. In an induction proof of the power rule, which expression results from differentiating xnxx^n x using the induction hypothesis (xn)′=nxn−1(x^n)'=nx^{n-1}?Answer: x(nxn−1)+xnx(nx^{n-1})+x^n

Based on the Wikipedia article Product rule, licensed CC BY-SA 4.0. Adapted into a summary, flashcards and quizzes.

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