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Mathematics LibreTexts

The Product and Quotient Rules

  • Page ID
    626
  •  

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    Theorem: The Product Rule

    Let \(f\) and \(g\) be differentiable functions. Then

    \[ \left[f(x) \, g(x)\right] ' = f(x)\, g '(x) + f '(x) \,g(x) \]

    Proof

    We have 

    prodqu1.gif

    Example \(\PageIndex{1}\)

    Find

    \[\dfrac{d}{dx } (2 - x^2)(x^4 - 5) \nonumber\]

    Solution:

    Here

    \[ f(x) = 2 - x^2  \nonumber\]

    and

    \[  g(x) = x^4 - 5 \nonumber\]

    The product rule gives

    \[\dfrac{d}{dx } (2 - x^2)(x^4 - 5) = (2 - x^2)(4x^3) + (-2x)(x^4 - 5) \nonumber\]

    The Quotient Rule

    Remember the poem

    "lo d hi minus hi d lo square the bottom and away you go"

    This poem is the mnemonic for the taking the derivative of a quotient.

    Theorem: The Quotient Rule

    Let f and g be differentiable functions. Then

    \[ \dfrac{d}{dx} \dfrac{f(x)}{g(x)} = \dfrac{g(x)\, f'(x) - f(x) \, g'(x)}{g(x)^2} \label{quot} \]

    Example \(\PageIndex{2}\):

    Find \(y'\) if

    \[ y' =\dfrac{2x - 1}{x + 1} \nonumber\]

    Solution

    Here

    \[ f(x) = 2x - 1 \nonumber\]

    and

    \[ g(x) = x + 1 \nonumber\]

    The quotient rule (Equation \ref{quot}) gives

    \[\begin{align*} \dfrac{ (x + 1)(2) - (2x - 1)(1)}{ (x + 1)2} &= \dfrac{2x + 2 - 2x + 1}{(x + 1)2} \\[5pt] &= \dfrac{3}{ (x + 1)2} \end{align*}\]

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