Question

`r = \frac{a { }^(2) + 2a - 7 }{ √(a) }`

differentiate this using the quotient rule


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Answer 1

`(dr)/(da) = \frac{3a^2 + 2a + 7}{2a^{(3)/(2) }}`

General Formulas and Concepts:

Pre-Algebra

• Distributive Property

Algebra I

• Expand by FOIL (First Outside Inside Last)
• Terms/Coefficients/Degrees

Algebra II

• Exponential Rule:
  `x^(-m)= (1)/(x^m)`
• Exponential Rule:
  `√(x) = x^{(1)/(2) }`

Calculus

Derivatives

Derivative Notation

The derivative of a constant is equal to 0

Basic Power Rule:

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Quotient Rule:
`\displaystyle (d)/(dx) [(f(x))/(g(x)) ]=(g(x)f'(x)-g'(x)f(x))/(g^2(x))`

Explanation:

Step 1: Define

`r = (a^2+2a-7)/(√(a))`

Step 2: Rewrite

`r = \frac{a^2+2a-7}{a^{(1)/(2) }}`

Step 3: Differentiate

1. Quotient Rule [Basic Power:
   `(dr)/(da) = \frac{a^{(1)/(2) }(2a^(2-1) + 2a^(1-1)) - (1)/(2)a^{(1)/(2) - 1 }(a^2 + 2a + 7)}{(a^{(1)/(2) })^2}`
2. Simplify:
   `(dr)/(da) = \frac{a^{(1)/(2) }(2a + 2) - (1)/(2)a^{-(1)/(2)}(a^2 + 2a + 7)}{a}`
3. Simplify:
   `(dr)/(da) = (2)/(2) \cdot \frac{a^{(1)/(2) }(2a + 2) - (1)/(2)a^{-(1)/(2)}(a^2 + 2a + 7)}{a}`
4. Multiply:
   `(dr)/(da) = \frac{2a^{(1)/(2) }(2a + 2) - a^{-(1)/(2)}(a^2 + 2a + 7)}{2a}`
5. Factor:
   `(dr)/(da) = \frac{a^{-(1)/(2)}[2a(2a + 2) - (a^2 + 2a + 7)]}{2a}`
6. [Brackets] Distribute:
   `(dr)/(da) = \frac{a^{-(1)/(2)}[4a^2 + 4a - a^2 - 2a - 7]}{2a}`
7. [Brackets] Combine Like Terms:
   `(dr)/(da) = \frac{a^{-(1)/(2)}[3a^2 + 2a - 7]}{2a}`
8. Simplify:
   `(dr)/(da) = \frac{3a^2 + 2a + 7}{2a^{(3)/(2) }}`