Mathematics, 09.03.2021 09:00 jimena15

Evaluate f′ (1) and f′′ (1): = x√x

3√ 5

Answers: 3

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Answers

Answer from: wardlawshaliyah

$\displaystyle f'(1) = \frac{3}{2}$

$\displaystyle f''(1) = \frac{3}{4}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

BracketsParenthesisExponentsMultiplicationDivisionAdditionSubtractionLeft to Right

Algebra II

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 Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

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

Derivative Rule [Quotient Rule]: $\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}$

Step-by-step explanation:

Step 1: Define

f(x) = x√x

f'(1) is x = 1 for 1st derivative

f''(1) is x = 1 for 2nd derivative

Step 2: Differentiate

[1st Derivative] Product Rule: $\displaystyle f'(x) = \frac{d}{dx}[x]\sqrt{x} + x\frac{d}{dx}[\sqrt{x}]$ [1st Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle f'(x) = \frac{d}{dx}[x]\sqrt{x} + x\frac{d}{dx}[x^{\frac{1}{2}}]$ [1st Derivative] Basic Power Rule: $\displaystyle f'(x) = (1 \cdot x^{1 - 1})\sqrt{x} + x(\frac{1}{2}x^{\frac{1}{2}-1})$ [1st Derivative] Simply Exponents: $\displaystyle f'(x) = (1 \cdot x^0)\sqrt{x} + x(\frac{1}{2}x^{\frac{-1}{2}})$ [1st Derivative] Simplify: $\displaystyle f'(x) = \sqrt{x} + x(\frac{1}{2}x^{\frac{-1}{2}})$ [1st Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle f'(x) = \sqrt{x} + x(\frac{1}{2x^{\frac{1}{2}}})$ [1st Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle f'(x) = \sqrt{x} + x(\frac{1}{2\sqrt{x}})$ [1st Derivative] Multiply: $\displaystyle f'(x) = \sqrt{x} + \frac{x}{2\sqrt{x}}$ [2nd Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle f'(x) = x^{\frac{1}{2}} + \frac{x}{2x^{\frac{1}{2}}}$ [2nd Derivative] Basic Power Rule/Quotient Rule [Derivative Property]: $\displaystyle f''(x) = \frac{1}{2}x^{\frac{1}{2} - 1} + \frac{\frac{d}{dx}[x](2x^{\frac{1}{2}}) - x\frac{d}{dx}[2x^{\frac{1}{2}}]}{(2x^{\frac{1}{2}})^2}$ [2nd Derivative] Simplify/Evaluate Exponents: $\displaystyle f''(x) = \frac{1}{2}x^{\frac{-1}{2}} + \frac{\frac{d}{dx}[x](2x^{\frac{1}{2}}) - x\frac{d}{dx}[2x^{\frac{1}{2}}]}{4x}$ [2nd Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{\frac{d}{dx}[x](2x^{\frac{1}{2}}) - x\frac{d}{dx}[2x^{\frac{1}{2}}]}{4x}$ [2nd Derivative] Basic Power Rule: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{(1 \cdot x^{1 - 1})(2x^{\frac{1}{2}}) - x(\frac{1}{2} \cdot 2x^{\frac{1}{2} - 1})}{4x}$ [2nd Derivative] Simply Exponents: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{(1 \cdot x^0)(2x^{\frac{1}{2}}) - x(\frac{1}{2} \cdot 2x^{\frac{-1}{2}})}{4x}$ [2nd Derivative] Simplify: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - x(\frac{1}{2} \cdot 2x^{\frac{-1}{2}})}{4x}$ [2nd Derivative] Multiply: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - x(x^{\frac{-1}{2}})}{4x}$ [2nd Derivative] Rewrite [Exponential Rule - Rewrite]: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - x(\frac{1}{x^{\frac{1}{2}}})}{4x}$ [2nd Derivative] Multiply: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{2x^{\frac{1}{2}} - \frac{x}{x^{\frac{1}{2}}}}{4x}$ [2nd Derivative] Simplify: $\displaystyle f''(x) = \frac{1}{2x^{\frac{1}{2}}} + \frac{x^{\frac{1}{2}}}{4x}$ [2nd Derivative] Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle f''(x) = \frac{1}{2\sqrt{x}} + \frac{\sqrt{x}}{4x}$

Step 3: Evaluate

[1st Derivative] Substitute in x: $\displaystyle f'(1) = \sqrt{1} + \frac{1}{2\sqrt{1}}$ [1st Derivative] Evaluate Roots: $\displaystyle f'(1) = 1 + \frac{1}{2(1)}$ [1st Derivative] Multiply: $\displaystyle f'(1) = 1 + \frac{1}{2}$ [1st Derivative] Add: $\displaystyle f'(1) = \frac{3}{2}$ [2nd Derivative] Substitute in x: $\displaystyle f''(1) = \frac{1}{2\sqrt{1}} + \frac{\sqrt{1}}{4(1)}$ [2nd Derivative] Evaluate Roots: $\displaystyle f''(1) = \frac{1}{2(1)} + \frac{1}{4(1)}$ [2nd Derivative] Multiply: $\displaystyle f''(1) = \frac{1}{2} + \frac{1}{4}$ [2nd Derivative] Add: $\displaystyle f''(1) = \frac{3}{4}$

Answer from: Quest

the answer is d

step-by-step explanation:

y+3=-4(x-2)

y+3=-4x+8

-3 -3

y=-4x+5

Which of the following equations is equivalent to y+3=-4(x-2)?

Answer from: Quest

B) greg: 30 points , tremaine 38 points

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