Answer :
Certainly! Let's work through this step-by-step to determine the \( p+q \)-th derivative of the function \( y = x^p (1 + x)^q \).
First, let's define the given functions:
[tex]\[ u = x^p \][/tex]
[tex]\[ v = (1 + x)^q \][/tex]
We will use the formula for the \( n \)-th derivative of the product of \( u \) and \( v \):
[tex]\[ y^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(n-k)} v^{(k)} \][/tex]
Here, we are interested in \( y^{(p+q)} \), so \( n = p + q \).
### Step-by-Step Calculation
1. Determine \( u \) and its derivatives:
[tex]\[ u = x^p \][/tex]
[tex]\[ u^{(n)} = \frac{d^n}{dx^n} (x^p) \][/tex]
For \( u = x^p \), the \( n \)-th derivative \( u^{(n)} \) is zero for \( n > p \). For \( n \leq p \):
[tex]\[ u^{(n)} = \frac{d^n}{dx^n} (x^p) = \frac{p!}{(p-n)!} x^{p-n} \][/tex]
2. Determine \( v \) and its derivatives:
[tex]\[ v = (1 + x)^q \][/tex]
[tex]\[ v^{(n)} = \frac{d^n}{dx^n} ((1 + x)^q) \][/tex]
For \( v = (1 + x)^q \), the \( n \)-th derivative \( v^{(n)} \) is given by:
[tex]\[ v^{(n)} = \frac{q!}{(q-n)!} (1+x)^{q-n} \][/tex]
3. Apply the product rule formula:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p+q} \binom{p+q}{k} u^{(p+q-k)} v^{(k)} \][/tex]
Now, let’s break down the terms for \( u^{(p+q-k)} \) and \( v^{(k)} \):
- For \( k > p \), \( u^{(p+q-k)} = 0 \) because the derivative \( u^{(n)} \) is zero when \( n > p \).
- For \( k \leq p \), \( u^{(p+q-k)} = \frac{p!}{(p - (p+q-k))!} x^{p - (p+q-k)} = \frac{p!}{(q-k)!} x^{p+q-k} \).
So, when we sum up these terms:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p} \binom{p+q}{k} \frac{p!}{(q-k)!} x^{p+q-k} \frac{q!}{(q-k)!} (1 + x)^{q-k} \][/tex]
Combining constants and simplifying:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p} \binom{p+q}{k} \frac{p! q!}{(q-k)!(q-k)!} x^{p+q-k} (1 + x)^{q-k} \][/tex]
The general form can be complex to express in a simplified term without specific values for \( p \) and \( q \). However, this approach gives us the detailed step-by-step method used to calculate higher-order derivatives.
### Summary
To calculate \( y^{(p+q)} \) for \( y = x^p (1 + x)^q \), we use:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p} \binom{p+q}{k} \frac{p! q!}{(q-k)!(q-k)!} x^{p+q-k} (1 + x)^{q-k} \][/tex]
This involves calculating the derivatives of the individual functions and then using the product rule to sum up the appropriate terms.
First, let's define the given functions:
[tex]\[ u = x^p \][/tex]
[tex]\[ v = (1 + x)^q \][/tex]
We will use the formula for the \( n \)-th derivative of the product of \( u \) and \( v \):
[tex]\[ y^{(n)} = \sum_{k=0}^{n} \binom{n}{k} u^{(n-k)} v^{(k)} \][/tex]
Here, we are interested in \( y^{(p+q)} \), so \( n = p + q \).
### Step-by-Step Calculation
1. Determine \( u \) and its derivatives:
[tex]\[ u = x^p \][/tex]
[tex]\[ u^{(n)} = \frac{d^n}{dx^n} (x^p) \][/tex]
For \( u = x^p \), the \( n \)-th derivative \( u^{(n)} \) is zero for \( n > p \). For \( n \leq p \):
[tex]\[ u^{(n)} = \frac{d^n}{dx^n} (x^p) = \frac{p!}{(p-n)!} x^{p-n} \][/tex]
2. Determine \( v \) and its derivatives:
[tex]\[ v = (1 + x)^q \][/tex]
[tex]\[ v^{(n)} = \frac{d^n}{dx^n} ((1 + x)^q) \][/tex]
For \( v = (1 + x)^q \), the \( n \)-th derivative \( v^{(n)} \) is given by:
[tex]\[ v^{(n)} = \frac{q!}{(q-n)!} (1+x)^{q-n} \][/tex]
3. Apply the product rule formula:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p+q} \binom{p+q}{k} u^{(p+q-k)} v^{(k)} \][/tex]
Now, let’s break down the terms for \( u^{(p+q-k)} \) and \( v^{(k)} \):
- For \( k > p \), \( u^{(p+q-k)} = 0 \) because the derivative \( u^{(n)} \) is zero when \( n > p \).
- For \( k \leq p \), \( u^{(p+q-k)} = \frac{p!}{(p - (p+q-k))!} x^{p - (p+q-k)} = \frac{p!}{(q-k)!} x^{p+q-k} \).
So, when we sum up these terms:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p} \binom{p+q}{k} \frac{p!}{(q-k)!} x^{p+q-k} \frac{q!}{(q-k)!} (1 + x)^{q-k} \][/tex]
Combining constants and simplifying:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p} \binom{p+q}{k} \frac{p! q!}{(q-k)!(q-k)!} x^{p+q-k} (1 + x)^{q-k} \][/tex]
The general form can be complex to express in a simplified term without specific values for \( p \) and \( q \). However, this approach gives us the detailed step-by-step method used to calculate higher-order derivatives.
### Summary
To calculate \( y^{(p+q)} \) for \( y = x^p (1 + x)^q \), we use:
[tex]\[ y^{(p+q)} = \sum_{k=0}^{p} \binom{p+q}{k} \frac{p! q!}{(q-k)!(q-k)!} x^{p+q-k} (1 + x)^{q-k} \][/tex]
This involves calculating the derivatives of the individual functions and then using the product rule to sum up the appropriate terms.
