find from first principle the derivative of 3x+5/√x July 24, 2021 by Sapo find from first principle the derivative of 3x+5/√x
Answer: [tex]\displaystyle \frac{d}{dx} = \frac{3x – 5}{2x^\bigg{\frac{3}{2}}}[/tex] General Formulas and Concepts: Algebra I Exponential Rule [Powering]: [tex]\displaystyle (b^m)^n = b^{m \cdot n}[/tex] Exponential Rule [Rewrite]: [tex]\displaystyle b^{-m} = \frac{1}{b^m}[/tex] Exponential Rule [Root Rewrite]: [tex]\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}[/tex] Calculus Derivatives Derivative Notation Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex] Basic Power Rule: f(x) = cxⁿ f’(x) = c·nxⁿ⁻¹ Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex] Step-by-step explanation: Step 1: Define Identify [tex]\displaystyle \frac{3x + 5}{\sqrt{x}}[/tex] Step 2: Differentiate Rewrite [Exponential Rule – Root Rewrite]: [tex]\displaystyle \frac{3x + 5}{x^\bigg{\frac{1}{2}}}[/tex] Quotient Rule: [tex]\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}[/tex] Simplify [Exponential Rule – Powering]: [tex]\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}[/tex] Basic Power Rule [Derivative Property – Addition/Subtraction]: [tex]\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}[/tex] Simplify: [tex]\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} – (\frac{1}{2}x^\bigg{\frac{-1}{2}})(3x + 5)}{x}[/tex] Rewrite [Exponential Rule – Rewrite]: [tex]\displaystyle \frac{d}{dx} = \frac{3x^\bigg{\frac{1}{2}} – (\frac{1}{2x^{\frac{1}{2}}})(3x + 5)}{x}[/tex] Rewrite [Exponential Rule – Root Rewrite]: [tex]\displaystyle \frac{d}{dx} = \frac{3\sqrt{x} – (\frac{1}{2\sqrt{x}})(3x + 5)}{x}[/tex] Simplify [Rationalize]: [tex]\displaystyle \frac{d}{dx} = \frac{3x – 5}{2x^\bigg{\frac{3}{2}}}[/tex] Topic: AP Calculus AB/BC (Calculus I/I + II) Unit: Derivatives Book: College Calculus 10e Reply
Answer:
[tex]\displaystyle \frac{d}{dx} = \frac{3x – 5}{2x^\bigg{\frac{3}{2}}}[/tex]
General Formulas and Concepts:
Algebra I
Calculus
Derivatives
Derivative Notation
Derivative Property [Addition/Subtraction]: [tex]\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)][/tex]
Basic Power Rule:
Derivative Rule [Quotient Rule]: [tex]\displaystyle \frac{d}{dx} [\frac{f(x)}{g(x)} ]=\frac{g(x)f'(x)-g'(x)f(x)}{g^2(x)}[/tex]
Step-by-step explanation:
Step 1: Define
Identify
[tex]\displaystyle \frac{3x + 5}{\sqrt{x}}[/tex]
Step 2: Differentiate
Topic: AP Calculus AB/BC (Calculus I/I + II)
Unit: Derivatives
Book: College Calculus 10e