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Mathematics LibreTexts

The Chain Rule

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    625

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    Our goal is to differentiate functions such as

    \[ y = (3x + 1)^{10} \]

    The last thing that we would want to do is FOIL this out ten times. We now look for a better way.

    Definition: The Chain Rule

    If \( y = y(u) \) is a function of \(u\), and \(u = u(x) \) is a function of \(x\) then \[\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \]

    In our example we have

    \[ y = u^{10} \]

    and

    \[ u = 3x + 1 \]

    so that

    \[\dfrac{dy}{dx} = \dfrac{dy}{du} \dfrac{du}{dx} \]

    = (10u9)(3) = 30(3x+1)9

    Proof: Chain Rule

    Recall an alternate definition of the derivative:

    chain.1.gif
    Example 2

    Find f '(x) if

    \[ f(x) = (x^4 - 3x^3 + x)^5 \]

    Solution

    Here

    \[ f(u) = u^20 \]

    and

    \[ u(x) = x^3 - x + 1 \]

    So that the derivative is

    \[ (20u^{19})(3x^2 - 1) = \left[20(x^3 - x + 1)^{19}\right](3x^2 - 1) \]
    Example 3

    Find f '(x) if

    \[ f(x) = (x^3 - x + 1)^{20} \]

    Solution

    Here

    \[ f(u) = u^5 \]

    and

    \[ u(x) = x^4 - 3x^3 + x \]

    So that the derivative is

    (5u4)(4x3 - 9x2 + 1) = [5(x4 - 3x3 + x)4](4x3 - 9x2 + 1)
    Example 4

    Find f '(x) if

    \[ f(x) = (1 - x)^9 (1-x^2)^4 \]

    Solution

    Here we need both the product and the chain rule. First the product rule

    f '(x) = [(1 - x)9][(1 - x2)4] ' + [(1 - x)9]' [(1 - x2)4]

    Now compute

    [(1 - x2)4]' = [4(1 - x2)3](-2x)

    and

    [(1 - x)9]' = [9(1 - x)8](-1)

    Putting this all together gives

    f '(x) = [(1 - x)9][4(1 - x2)3](-2x) - [9(1 - x)8] [(1 - x2)4]
    Example 5

    Find f '(x) if

    \[ f(x)= \dfrac{ (x^3 + 4x - 3)^7}{ (2x - 1)^3} \]

    Solution

    Here we need both the quotient and the chain rule.

    \[ f'(x) = \dfrac{(2x - 1)^3\left[(x^3 + 4x - 3)^7\right]' - (x^3 + 4x - 3)^7 \left[(2x - 1)^3\right]'}{(2x - 1)^6} \]

    We first compute

    [(x3 + 4x - 3)7]' = [7(x3 + 4x - 3)6](3x2 + 4)

    and

    [(2x - 1)3]' = [3(2x - 1)2](2)

    Putting this all together gives

    \[ f](x) = \dfrac{7(2x - 1)^3(x^3 + 4x - 3)^6(3x^2 + 4) + 6(x^3 + 4x - 3)^7 (2x - 1)^2}{ (2x - 1)^6 } \]

    \[ f(x)= \dfrac{x^2(5 - x^3)^4 }{3 - x} \]

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