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

Sagot :

leena

Hi there!

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

[tex]y = \sqrt{x} * sec^{-1}(-\sqrt{x}})[/tex]

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)

[tex]\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}}[/tex]

Simplify:

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

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

Space

Answer:

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

General Formulas and Concepts:

Algebra I

  • 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

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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

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

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

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle y = \sqrt{x}sec^{-1}(\sqrt{x})[/tex]

Step 2: Differentiate

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

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

Unit: Derivatives

Book: College Calculus 10e

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