## find from first principle the derivative of 3x+5/√x​

Question

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

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2 years 2021-07-24T02:09:10+00:00 1 Answers 7 views 0

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

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:

1. f(x) = cxⁿ
2. 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

1. Rewrite [Exponential Rule – Root Rewrite]:                                                     $$\displaystyle \frac{3x + 5}{x^\bigg{\frac{1}{2}}}$$
2. Quotient Rule:                                                                                                   $$\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})\frac{d}{dx}[3x + 5] – \frac{d}{dx}[x^\bigg{\frac{1}{2}}](3x + 5)}{(x^\bigg{\frac{1}{2}})^2}$$
3. Simplify [Exponential Rule – Powering]:                                                          $$\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})\frac{d}{dx}[3x + 5] – \frac{d}{dx}[x^\bigg{\frac{1}{2}}](3x + 5)}{x}$$
4. Basic Power Rule [Derivative Property – Addition/Subtraction]:                   $$\displaystyle \frac{d}{dx} = \frac{(x^\bigg{\frac{1}{2}})(3x^{1 – 1} + 0) – (\frac{1}{2}x^\bigg{\frac{1}{2} – 1})(3x + 5)}{x}$$
5. Simplify:                                                                                                             $$\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} – (\frac{1}{2}x^\bigg{\frac{-1}{2}})(3x + 5)}{x}$$
6. Rewrite [Exponential Rule – Rewrite]:                                                              $$\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} – (\frac{1}{2x^{\frac{1}{2}}})(3x + 5)}{x}$$
7. Rewrite [Exponential Rule – Root Rewrite]:                                                     $$\displaystyle \frac{d}{dx} = \frac{3\sqrt{x} – (\frac{1}{2\sqrt{x}})(3x + 5)}{x}$$
8. 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