35.7k views
5 votes
Find dy/dx of the function y = √x sec*-1 (√x)​

2 Answers

3 votes

Hi there!


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


y = √(x) * sec^(-1)(-√(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)


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

Simplify:


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


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

User Pasquers
by
7.4k points
1 vote

Answer:


\displaystyle y' = (arcsec(√(x)))/(2√(x)) + (1)/(2|√(x)|√(x - 1))

General Formulas and Concepts:

Algebra I

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

Calculus

Derivatives

Derivative Notation

Basic Power Rule:

  • f(x) = cxⁿ
  • f’(x) = c·nxⁿ⁻¹

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

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

Arctrig Derivative:
\displaystyle (d)/(dx)[arcsec(u)] = (u')/(|u|√(u^2 - 1))

Explanation:

Step 1: Define

Identify


\displaystyle y = √(x)sec^(-1)(√(x))

Step 2: Differentiate

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

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

Unit: Derivatives

Book: College Calculus 10e

User Riley Hun
by
8.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories