Question

Find the first derivative?

Answer 1

`\displaystyle y' = (-1)/((x - 2)^2)`

General Formulas and Concepts:

Calculus

Differentiation

• Derivatives
• Derivative Notation

Derivative Property [Multiplied Constant]:
`\displaystyle (d)/(dx) [cf(x)] = c \cdot f'(x)`

Derivative Property [Addition/Subtraction]:
`\displaystyle (d)/(dx)[f(x) + g(x)] = (d)/(dx)[f(x)] + (d)/(dx)[g(x)]`

Basic Power Rule:

1. f(x) = cxⁿ
2. f’(x) = c·nxⁿ⁻¹

Derivative Rule [Quotient Rule]:
`\displaystyle (d)/(dx) [(f(x))/(g(x)) ]=(g(x)f'(x)-g'(x)f(x))/(g^2(x))`

Derivative Rule [Chain Rule]:
`\displaystyle (d)/(dx)[f(g(x))] =f'(g(x)) \cdot g'(x)`

Explanation:

Step 1: Define

Identify

`\displaystyle y = (1)/(x - 2) - 2`

Step 2: Differentiate

1. Derivative Rule [Quotient Rule]:
   `\displaystyle y' = (1'(x - 2) - 1(x - 2)')/((x - 2)^2)`
2. Basic Power Rule [Derivative Property - Addition/Subtraction]:
   `\displaystyle y' = (-1(1))/((x - 2)^2)`
3. Simplify:
   `\displaystyle y' = (-1)/((x - 2)^2)`

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

Unit: Differentiation

Answer 2

y = 1/(x - 2) - 3

y' = d(x - 2)^-1 / dx The three is a constant and will disappear.

y' = d(x - 2)^-1 -1 d(x - 2)/dx

y' = -1*(x - 2)^-2 * 1

The minus comes from bringing down the original power down and reducing the power by one of the question. The 1 at the end comes from differentiating what is inside the brackets. It is called the chain rule

d(x - 2) / dx = d(x) / dx which is 1. The two, being a constant, disappears. The final answer is

y' = - (x - 2)^-2