Problem 32 - Advanced Engineering Mathematics

Question

Find the first partial derivatives of the given function. $$ G(p, q, r, s)=\left(p^{2} q^{3}\right)^{r^{4} s^{5}} $$

Step-by-Step Solution

Answer

The first partial derivatives are found using power and chain rules for each variable.

Step 1: Identify the Function and Variables

The function given is \( G(p, q, r, s)=ig(p^{2} q^{3}\big)^{r^{4} s^{5}} \). The variables in the function are \( p, q, r, \text{ and } s \). We need to find the first partial derivatives with respect to each of these variables.

Step 2: Simplify the Expression

Notice that the expression inside the function can be rewritten for easier differentiation: \( G(p, q, r, s) = (p^{2})^{r^{4}s^{5}} \times (q^{3})^{r^{4}s^{5}} = p^{2r^{4}s^{5}} \times q^{3r^{4}s^{5}} \). This simplifies our differentiation process.

Step 3: Differentiate with Respect to p

For partial differentiation with respect to \( p \), treat \( q, r, s \) as constants. Use the power rule: \( \frac{\partial G}{\partial p} = 2r^{4}s^{5}p^{2r^{4}s^{5}-1} \times q^{3r^{4}s^{5}} \).

Step 4: Differentiate with Respect to q

For partial differentiation with respect to \( q \), treat \( p, r, s \) as constants. Use the power rule again: \( \frac{\partial G}{\partial q} = 3r^{4}s^{5}q^{3r^{4}s^{5}-1} \times p^{2r^{4}s^{5}} \).

Step 5: Differentiate with Respect to r

For partial differentiation with respect to \( r \), treat \( p, q, s \) as constants. Use the chain rule: \( \frac{\partial G}{\partial r} = G(p, q, r, s) \times \ln(p^{2} q^{3}) \times 4r^{3}s^{5} \).

Step 6: Differentiate with Respect to s

For partial differentiation with respect to \( s \), treat \( p, q, r \) as constants. Use the chain rule similarly: \( \frac{\partial G}{\partial s} = G(p, q, r, s) \times \ln(p^{2} q^{3}) \times 5r^{4}s^{4} \).

Key Concepts

Differentiation

Differentiation is a fundamental concept in calculus that allows us to find how a function changes with respect to its variables. When we differentiate a function, we calculate its derivative, which is the rate of change of the function with respect to one of its variables. In the context of multivariable calculus, we often deal with partial derivatives. A partial derivative is the derivative of a function with respect to just one of its variables, while keeping all other variables constant.
For instance, when finding the partial derivative of a function \( G(p, q, r, s) \) with respect to \( p \), we treat \( q, r, \text{ and } s \) as constants. This allows us to understand how changes in \( p \) alone will affect the function \( G \). Differentiation provides the tools needed to analyze these relationships.

Power Rule

The power rule is a key technique in differentiation. It is used when a function is expressed as a term raised to a power. The rule states that if you have a function \( x^n \), its derivative is determined by multiplying the exponent \( n \) by the original function raised to the power of \( n-1 \).
In mathematical terms, the power rule is expressed as:

• For \( f(x) = x^n \), the derivative \( f'(x) = n \cdot x^{n-1} \).

This rule simplifies differentiation of polynomial functions.
In the original problem, the function \( G(p, q, r, s) \) was simplified using the power rule when calculating the partial derivatives with respect to \( p \) and \( q \). Differentiating partially with respect to a particular variable involves applying the power rule to that specific variable while treating others as constants.

Chain Rule

The chain rule is another vital tool in calculus, particularly useful for handling composite functions. It helps us find the derivative of a function composed of other functions. The chain rule states that the derivative of a composite function \( f(g(x)) \) is found by taking the derivative of the outer function \( f \) evaluated at \( g(x) \), multiplied by the derivative of the inner function \( g(x) \).
Expressed mathematically, it looks like this:

• If \( h(x) = f(g(x)) \), then \( h'(x) = f'(g(x)) \cdot g'(x) \).

This concept is particularly useful when differentiating more complex expressions, such as those with nested functions.
In the problem provided, the chain rule was applied when finding partial derivatives with respect to \( r \) and \( s \). Here, the chain rule facilitated the differentiation by breaking down the composite nature of the function \( G(p, q, r, s) \).

Multivariable Calculus

Multivariable calculus expands on the principles of single-variable calculus to functions of multiple variables. This field of calculus deals with functions that have more than one input, allowing us to model and analyze real-world phenomena with multiple influencing factors.
Key operations in multivariable calculus include partial differentiation, multiple integrals, and directional derivatives. Understanding these concepts enables us to work with functions in higher dimensions than the typical one-variable scenarios.
In the provided exercise, the function \( G(p, q, r, s) \) depends on four variables \( p, q, r, \text{ and } s \). By finding the partial derivatives of this function, we can see how changes in any specific variable affect the overall function. Multivariable calculus, therefore, provides the framework for exploring these types of relationships in a rigorous mathematical way. This branch of calculus is essential in fields like engineering, physics, and economics, where multiple factors often influence outcomes simultaneously.