Determine the first and second derivative given the following functions. write the solution of each answers

Question

1. y= 1/x

Answer: -1/x² , 2/x³

2. y = 1/x (x² - 3x)²

Answer: 3x²-12x+9, 6x-12

Answer 1

1a. Find the first derivative of y = 1/x

First, let's rewrite 1/x using a negative exponent:

y = 1/x

y = x⁻¹

Next, find the derivative y' using the power rule:

y' = (-1) y⁻²

y' = −x⁻²

y' = −1/x²

1b. Find the second derivative of y = 1/x

Let's take the derivative of the result we found in part 1a (the second derivative is the derivative of the first derivative).

y' = −x⁻²

y'' = −(−2)x⁻³

y'' = 2x⁻³

y'' = 2/x³

2a. Find the first derivative of y = 1/x * (x² − 3x)²

First, let's try and simplify the right side of the equation as much as possible.

We can factor an x from the expression inside the parentheses

y = 1/x * (x * (x − 3))²

Then, distribute the square over the terms in the outer parentheses

y = 1/x * (x)² (x − 3)²

y = 1/x * x² (x − 3)²

The first two terms, 1/x and x² cancel to give:

y = x (x − 3)²

Next, foil the (x − 3)² term

y = x * (x² − 6x + 9)

Next, distribute the x over the terms in the parentheses

y = x³ − 6x² + 9x

Now, apply the power rule to take the derivative

y' = 3x² − (2)*6x + 9

y' = 3x² − 12x + 9

2a. Find the first derivative of y = 1/x * (x² − 3x)²

Let's take the derivative of the result we found in part 2a (the second derivative is the derivative of the first derivative).

y' = 3x² − 12x + 9

y'' = (2)*3x − 12

y'' = 6x − 12