Find dy/dx of the function y = √x sec*-1 (√x)​

Question

Find dy/dx of the function y = √x sec*-1 (√x)​

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Thanh Thu 3 years 2021-07-30T18:30:54+00:00 2 Answers 9 views 0

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  1. Thunguyet
    0
    2021-07-30T18:31:54+00:00
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    Answer:

    \displaystyle y' = \frac{arcsec(\sqrt{x})}{2\sqrt{x}} + \frac{1}{2|\sqrt{x}|\sqrt{x - 1}}

    General Formulas and Concepts:

    Algebra I

    • Exponential Rule [Rewrite]:                                                                           \displaystyle b^{-m} = \frac{1}{b^m}
    • Exponential Rule [Root Rewrite]:                                                                 \displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}

    Calculus

    Derivatives

    Derivative Notation

    Basic Power Rule:

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

    Derivative Rule [Product Rule]:                                                                             \displaystyle \frac{d}{dx} [f(x)g(x)]=f'(x)g(x) + g'(x)f(x)

    Derivative Rule [Chain Rule]:                                                                                 \displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)

    Arctrig Derivative:                                                                                                 \displaystyle \frac{d}{dx}[arcsec(u)] = \frac{u'}{|u|\sqrt{u^2 - 1}}

    Step-by-step explanation:

    Step 1: Define

    Identify

    \displaystyle y = \sqrt{x}sec^{-1}(\sqrt{x})

    Step 2: Differentiate

    1. Rewrite:                                                                                                         \displaystyle y = \sqrt{x}arcsec(\sqrt{x})
    2. Product Rule:                                                                                                \displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})]
    3. Chain Rule:                                                                                                     \displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[\sqrt{x}] \bigg]
    4. Rewrite [Exponential Rule – Root Rewrite]:                                                 \displaystyle y' = \frac{d}{dx}[x^\bigg{\frac{1}{2}}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[x^\bigg{\frac{1}{2}}] \bigg]
    5. Basic Power Rule:                                                                                         \displaystyle y' = \frac{1}{2}x^\bigg{\frac{1}{2} - 1}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{1}{2} - 1} \bigg]
    6. Simplify:                                                                                                         \displaystyle y' = \frac{1}{2}x^\bigg{\frac{-1}{2}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{-1}{2}} \bigg]
    7. Rewrite [Exponential Rule – Rewrite]:                                                           \displaystyle y' = \frac{1}{2x^\bigg{\frac{1}{2}}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2x^\bigg{\frac{1}{2}}} \bigg]
    8. Rewrite [Exponential Rule – Root Rewrite]:                                                 \displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2\sqrt{x}} \bigg]
    9. Arctrig Derivative:                                                                                         \displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{1}{|\sqrt{x}|\sqrt{(\sqrt{x})^2 - 1}} \cdot \frac{1}{2\sqrt{x}} \bigg]
    10. Simplify:                                                                                                         \displaystyle y' = \frac{arcsec(\sqrt{x})}{2\sqrt{x}} + \frac{1}{2|\sqrt{x}|\sqrt{x - 1}}

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

    Unit: Derivatives

    Book: College Calculus 10e

  2. Dau
    0
    2021-07-30T18:32:11+00:00
    Reply

    Hi there!

    \large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}

    y = \sqrt{x} * sec^{-1}(-\sqrt{x}})

    Use the chain rule and multiplication rules to solve:

    g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

    g(f(x)) = g'(f(x)) * ‘f(x))

    Thus:

    f(x) = √x

    g(x) = sec⁻¹ (√x)

    \frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}

    Simplify:

    \frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}

    \frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}

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