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

answer: the answer is one of

$f=\dfrac{1}{2},~\dfrac{1}{4},~\dfrac{3}{4},~\dfrac{1}{6},~\dfrac{5}{6},~\dfrac{1}{8},~\dfrac{3}{8},~\dfrac{5}{8},~\dfrac{7}{8}.$

step-by-step explanation: we are given to write a proper fraction where the whole is divided into an even number of equal parts less than ten, and an odd number of those parts exist.

an even number less than 10 may be one of 2, 4, 6 or 8. so, the whole may be one of these four evens.

now, an odd number of these parts must exist, so the numerator must be an odd number, one of 1, 3, 5 or 7. it cannot be 9 because then the fraction will become improper.

thus, our fraction may be one of the following

$f=\dfrac{1}{2},~\dfrac{1}{4},~\dfrac{3}{4},~\dfrac{1}{6},~\dfrac{5}{6},~\dfrac{1}{8},~\dfrac{3}{8},~\dfrac{5}{8},~\dfrac{7}{8}.$

Answer from: Quest

need points to ask question

step-by-step explanation:

Explain how you would find the measure of angle a. look at the image linked.

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Evaluate f′ (1) and f′′ (1): = x√x

3√ 5...

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