Question

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

Answer 1

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}}`

Answer 2

`\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