find from first principle the derivative of 3x+5/√x​

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find from first principle the derivative of 3x+5/√x​

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Sapo 2 years 2021-07-24T02:09:10+00:00 1 Answers 7 views 0

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  1. Thunguyet
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    2021-07-24T02:10:19+00:00
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    Answer:

    [tex]\displaystyle \frac{d}{dx} = \frac{3x – 5}{2x^\bigg{\frac{3}{2}}}[/tex]

    General Formulas and Concepts:

    Algebra I

    • Exponential Rule [Powering]:                                                                          [tex]\displaystyle (b^m)^n = b^{m \cdot n}[/tex]
    • Exponential Rule [Rewrite]:                                                                              [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex]
    • Exponential Rule [Root Rewrite]:                                                                     [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex]

    Calculus

    Derivatives

    Derivative Notation

    Derivative Property [Addition/Subtraction]:                                                            [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]

    Basic Power Rule:

    1. f(x) = cxⁿ
    2. f’(x) = c·nxⁿ⁻¹

    Derivative Rule [Quotient Rule]:                                                                               [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]

    Step-by-step explanation:

    Step 1: Define

    Identify

    [tex]\displaystyle \frac{3x + 5}{\sqrt{x}}[/tex]

    Step 2: Differentiate

    1. Rewrite [Exponential Rule – Root Rewrite]:                                                     [tex]\displaystyle \frac{3x + 5}{x^\bigg{\frac{1}{2}}}[/tex]
    2. Quotient Rule:                                                                                                   [tex]\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})\frac{d}{dx}[3x + 5] – \frac{d}{dx}[x^\bigg{\frac{1}{2}}](3x + 5)}{(x^\bigg{\frac{1}{2}})^2}[/tex]
    3. Simplify [Exponential Rule – Powering]:                                                          [tex]\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})\frac{d}{dx}[3x + 5] – \frac{d}{dx}[x^\bigg{\frac{1}{2}}](3x + 5)}{x}[/tex]
    4. Basic Power Rule [Derivative Property – Addition/Subtraction]:                   [tex]\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})(3x^{1 – 1} + 0) – (\frac{1}{2}x^\bigg{\frac{1}{2} – 1})(3x + 5)}{x}[/tex]
    5. Simplify:                                                                                                             [tex]\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} – (\frac{1}{2}x^\bigg{\frac{-1}{2}})(3x + 5)}{x}[/tex]
    6. Rewrite [Exponential Rule – Rewrite]:                                                              [tex]\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} – (\frac{1}{2x^{\frac{1}{2}}})(3x + 5)}{x}[/tex]
    7. Rewrite [Exponential Rule – Root Rewrite]:                                                     [tex]\displaystyle \frac{d}{dx} = \frac{3\sqrt{x} – (\frac{1}{2\sqrt{x}})(3x + 5)}{x}[/tex]
    8. Simplify [Rationalize]:                                                                                       [tex]\displaystyle \frac{d}{dx} = \frac{3x – 5}{2x^\bigg{\frac{3}{2}}}[/tex]

    Topic: AP Calculus AB/BC (Calculus I/I + II)

    Unit: Derivatives

    Book: College Calculus 10e

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