Mathematics, 09.07.2021 05:40 jessicaflower277

Find from first principle the derivative of 3x+5/√x

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Answer from: justyne2004

$\displaystyle \frac{d}{dx} = \frac{3x - 5}{2x^\bigg{\frac{3}{2}}}$

General Formulas and Concepts:

Algebra I

Exponential Rule [Powering]: $\displaystyle (b^m)^n = b^{m \cdot n}$ 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

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

Basic Power Rule:

f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹

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

Identify

$\displaystyle \frac{3x + 5}{\sqrt{x}}$

Step 2: Differentiate

Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle \frac{3x + 5}{x^\bigg{\frac{1}{2}}}$ Quotient Rule: $\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})\frac{d}{dx}[3x + 5] - \frac{d}{dx}[x^\bigg{\frac{1}{2}}](3x + 5)}{(x^\bigg{\frac{1}{2}})^2}$ Simplify [Exponential Rule - Powering]: $\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})\frac{d}{dx}[3x + 5] - \frac{d}{dx}[x^\bigg{\frac{1}{2}}](3x + 5)}{x}$ Basic Power Rule [Derivative Property - Addition/Subtraction]: $\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})(3x^{1 - 1} + 0) - (\frac{1}{2}x^\bigg{\frac{1}{2} - 1})(3x + 5)}{x}$ Simplify: $\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} - (\frac{1}{2}x^\bigg{\frac{-1}{2}})(3x + 5)}{x}$ Rewrite [Exponential Rule - Rewrite]: $\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} - (\frac{1}{2x^{\frac{1}{2}}})(3x + 5)}{x}$ Rewrite [Exponential Rule - Root Rewrite]: $\displaystyle \frac{d}{dx} = \frac{3\sqrt{x} - (\frac{1}{2\sqrt{x}})(3x + 5)}{x}$ Simplify [Rationalize]: $\displaystyle \frac{d}{dx} = \frac{3x - 5}{2x^\bigg{\frac{3}{2}}}$

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

Unit: Derivatives

Book: College Calculus 10e

Answer from: Quest

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step-by-step explanation:

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Given: efgh is a square find the length of fh.

Answer from: Quest

$\bf ~~~~~~~~~~~~\textit{distance between 2 points} \\\\ a(\stackrel{x_1}{5}~,~\stackrel{y_1}{4})\qquad b(\stackrel{x_2}{x}~,~\stackrel{y_2}{y})\qquad \qquad d = \sqrt{( x_2- x_1)^2 + ( y_2- y_1)^2} \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \stackrel{ab}{7}=\sqrt{(x-5)^2+(y-4)^2}~\hfill$