Answer :
To find the formula for the \( n \)th derivative \( y^{(n)} \) of the function \( y = \frac{1}{x + 1} \), we can follow a pattern by calculating the first few derivatives and identifying a general form.
Given:
[tex]\[ y = \frac{1}{x + 1} \][/tex]
Let's calculate the first few derivatives:
1. First derivative \( y' \):
[tex]\[ y' = \frac{d}{dx} \left( \frac{1}{x + 1} \right) \][/tex]
Using the chain rule, we get:
[tex]\[ y' = -\frac{1}{(x + 1)^2} \][/tex]
2. Second derivative \( y'' \):
[tex]\[ y'' = \frac{d}{dx} \left( -\frac{1}{(x + 1)^2} \right) \][/tex]
Using the chain rule again, we get:
[tex]\[ y'' = -\frac{d}{dx} \left( (x + 1)^{-2} \right) \][/tex]
[tex]\[ y'' = -(-2) \cdot (x + 1)^{-3} \][/tex]
[tex]\[ y'' = \frac{2}{(x + 1)^3} \][/tex]
3. Third derivative \( y''' \):
[tex]\[ y''' = \frac{d}{dx} \left( \frac{2}{(x + 1)^3} \right) \][/tex]
Using the chain rule:
[tex]\[ y''' = 2 \cdot \frac{d}{dx} \left( (x + 1)^{-3} \right) \][/tex]
[tex]\[ y''' = 2 \cdot (-3) \cdot (x + 1)^{-4} \][/tex]
[tex]\[ y''' = -\frac{6}{(x + 1)^4} \][/tex]
From the above calculations, we observe a pattern in the derivatives. Specifically, each derivative alternates in sign and has the general form:
[tex]\[ y^{(n)} = \frac{(-1)^n \cdot n!}{(x + 1)^{n + 1}} \][/tex]
where \( n! \) (n factorial) represents the product of all positive integers up to \( n \), and \( (x + 1)^{n + 1} \) is the term in the denominator raised to the power \( n + 1 \).
Therefore, the formula for the \( n \)th derivative \( y^{(n)} \) of the function \( y = \frac{1}{x + 1} \) is:
[tex]\[ y^{(n)} = \frac{(-1)^n \cdot n!}{(x + 1)^{n + 1}} \][/tex]
Given:
[tex]\[ y = \frac{1}{x + 1} \][/tex]
Let's calculate the first few derivatives:
1. First derivative \( y' \):
[tex]\[ y' = \frac{d}{dx} \left( \frac{1}{x + 1} \right) \][/tex]
Using the chain rule, we get:
[tex]\[ y' = -\frac{1}{(x + 1)^2} \][/tex]
2. Second derivative \( y'' \):
[tex]\[ y'' = \frac{d}{dx} \left( -\frac{1}{(x + 1)^2} \right) \][/tex]
Using the chain rule again, we get:
[tex]\[ y'' = -\frac{d}{dx} \left( (x + 1)^{-2} \right) \][/tex]
[tex]\[ y'' = -(-2) \cdot (x + 1)^{-3} \][/tex]
[tex]\[ y'' = \frac{2}{(x + 1)^3} \][/tex]
3. Third derivative \( y''' \):
[tex]\[ y''' = \frac{d}{dx} \left( \frac{2}{(x + 1)^3} \right) \][/tex]
Using the chain rule:
[tex]\[ y''' = 2 \cdot \frac{d}{dx} \left( (x + 1)^{-3} \right) \][/tex]
[tex]\[ y''' = 2 \cdot (-3) \cdot (x + 1)^{-4} \][/tex]
[tex]\[ y''' = -\frac{6}{(x + 1)^4} \][/tex]
From the above calculations, we observe a pattern in the derivatives. Specifically, each derivative alternates in sign and has the general form:
[tex]\[ y^{(n)} = \frac{(-1)^n \cdot n!}{(x + 1)^{n + 1}} \][/tex]
where \( n! \) (n factorial) represents the product of all positive integers up to \( n \), and \( (x + 1)^{n + 1} \) is the term in the denominator raised to the power \( n + 1 \).
Therefore, the formula for the \( n \)th derivative \( y^{(n)} \) of the function \( y = \frac{1}{x + 1} \) is:
[tex]\[ y^{(n)} = \frac{(-1)^n \cdot n!}{(x + 1)^{n + 1}} \][/tex]
