If the gradient of the tangent to
[tex]y = \sqrt{x} [/tex]
is
[tex] \frac{1}{6} [/tex]
at point A, find the coordinates of A.

Coordinates (x, y)
Functions
Function Notation
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

Derivative of a constant is 0

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 = \sqrt{x}[/tex]

[tex]\displaystyle y' = \frac{1}{6}[/tex]

Step 2: Differentiate

[Function] Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle y = x^{\frac{1}{2}}[/tex]
Basic Power Rule: [tex]\displaystyle y' = \frac{1}{2}x^{\frac{1}{2} - 1}[/tex]
Simplify: [tex]\displaystyle y' = \frac{1}{2}x^{-\frac{1}{2}}[/tex]
[Derivative] Rewrite [Exponential Rule - Rewrite]: [tex]\displaystyle y' = \frac{1}{2x^{\frac{1}{2}}}[/tex]
[Derivative] Rewrite [Exponential Rule - Root Rewrite]: [tex]\displaystyle y' = \frac{1}{2\sqrt{x}}[/tex]

Step 3: Solve

Find coordinates of A.

x-coordinate

Substitute in y' [Derivative]: [tex]\displaystyle \frac{1}{6} = \frac{1}{2\sqrt{x}}[/tex]
[Multiplication Property of Equality] Multiply 2 on both sides: [tex]\displaystyle \frac{1}{3} = \frac{1}{\sqrt{x}}[/tex]
[Multiplication Property of Equality] Cross-multiply: [tex]\displaystyle \sqrt{x} = 3[/tex]
[Equality Property] Square both sides: [tex]\displaystyle x = 9[/tex]

y-coordinate

∴ Coordinates of A is (9, 3).

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

Unit: Derivatives

Book: College Calculus 10e

If the gradient of the tangent to [tex]y = \sqrt{x} [/tex]is [tex] \frac{1}{6} [/tex]at point A, find the coordinates of A.​