Mathematics, 05.07.2021 21:00 middlegirlrule6848

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

Answers: 1

Show answers

Answers

Answer from: joho38

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:

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

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

Answer from: kimseokjin92

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

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) = cxⁿf’(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

Rewrite: $\displaystyle y = \sqrt{x}arcsec(\sqrt{x})$ Product Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})]$ Chain Rule: $\displaystyle y' = \frac{d}{dx}[\sqrt{x}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[\sqrt{x}] \bigg]$ Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{d}{dx}[x^\bigg{\frac{1}{2}}]arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{d}{dx}[x^\bigg{\frac{1}{2}}] \bigg]$ Basic Power Rule: $\displaystyle y' = \frac{1}{2}x^\bigg{\frac{1}{2} - 1}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{1}{2} - 1} \bigg]$ Simplify: $\displaystyle y' = \frac{1}{2}x^\bigg{\frac{-1}{2}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2}x^\bigg{\frac{-1}{2}} \bigg]$ Rewrite [Exponential Rule - Rewrite]: $\displaystyle y' = \frac{1}{2x^\bigg{\frac{1}{2}}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2x^\bigg{\frac{1}{2}}} \bigg]$ Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{d}{dx}[arcsec(\sqrt{x})] \cdot \frac{1}{2\sqrt{x}} \bigg]$ Arctrig Derivative: $\displaystyle y' = \frac{1}{2\sqrt{x}}arcsec(\sqrt{x}) + \bigg[ \sqrt{x}\frac{1}{|\sqrt{x}|\sqrt{(\sqrt{x})^2 - 1}} \cdot \frac{1}{2\sqrt{x}} \bigg]$ 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

Answer from: Quest

a: 1/6, j: 1/4

step-by-step explanation: you are welcome

Answer from: Quest

step-by-step explanation:

Another question on Mathematics

Mathematics, 21.06.2019 14:50

An assembly consists of two mechanical components. suppose that the probabilities that the first and second components meet specifications are 0.87 and 0.84. assume that the components are independent. determine the probability mass function of the number of components in the assembly that meet specifications. x

Answers: 1

Answer

Mathematics, 21.06.2019 17:30

Two customers went to a post office. first customer paid $12 for 14 and 5 envelopes the second paid 24.80 for 10 postcards and 15 envelopes. what is cost of each envelope?

Answers: 1

Answer

Mathematics, 21.06.2019 18:50

Determine if the function below is continuous. a. continuous b. not continuous, 1 hole c. not continuous, > 2 holes d. not continuous, 2 holes

Answers: 1

Answer

Mathematics, 21.06.2019 22:30

Convert (-3,0) to polar form. a. (3,0’) b. (-3,180’) c. (3,180’) d. (3,360’)

Answers: 1

Answer

You know the right answer?

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