Find
[tex] \frac{dy}{dx} [/tex]
if
[tex]y = (x + \sqrt{x} )^{2} [/tex]

Terms/Coefficients
Anything to the 0th power is 1
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

Basic Power Rule:

Derivative Rule [Chain Rule]: [tex]\displaystyle \frac{d}{dx}[f(g(x))] =f'(g(x)) \cdot g'(x)[/tex]

Step-by-step explanation:

Step 1: Define

Identify

[tex]\displaystyle y = (x + \sqrt{x})^2[/tex]

Step 2: Differentiate

Chain Rule: [tex]\displaystyle y' = 2(x + \sqrt{x})^{2 - 1} \cdot \frac{d}{dx}[x + \sqrt{x}][/tex]
Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle y' = 2(x + x^{\frac{1}{2}})^{2 - 1} \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}][/tex]
Simplify: [tex]\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot \frac{d}{dx}[x + x^{\frac{1}{2}}][/tex]
Basic Power Rule: [tex]\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 \cdot x^{1 - 1} + \frac{1}{2}x^{\frac{1}{2} - 1})[/tex]
Simplify: [tex]\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2}x^{-\frac{1}{2}})[/tex]
Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle y' = 2(x + x^{\frac{1}{2}}) \cdot (1 + \frac{1}{2x^{\frac{1}{2}}})[/tex]
Multiply: [tex]\displaystyle y' = 2[(x + x^{\frac{1}{2}}) + \frac{x + x^{\frac{1}{2}}}{2x^{\frac{1}{2}}}][/tex]
[Brackets] Add: [tex]\displaystyle y' = 2(\frac{2x + 3x^{\frac{1}{2}} + 1}{2})[/tex]
Multiply: [tex]\displaystyle y' = 2x + 3x^{\frac{1}{2}} + 1[/tex]
Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle y' = 2x + 3\sqrt{x} + 1[/tex]

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

Unit: Derivatives

Book: College Calculus 10e

Find [tex] \frac{dy}{dx} [/tex]if [tex]y = (x + \sqrt{x} )^{2} [/tex]​