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The Chain Rule is essential for computing partial derivatives, especially when dealing with composite functions. In the context of the problem, it helps us differentiate a function of multiple variables by breaking it down into simpler parts. The chain rule formula for a function \( u(x, y) \) in terms of \( r \) and \( s \) is:
\[ \frac{\partial u}{\partial r} = \frac{\partial u}{\partial x} \cdot \frac{\partial x}{\partial r} + \frac{\partial u}{\partial y} \cdot \frac{\partial y}{\partial r} \]
This means we need to find the partial derivatives of \( u \) with respect to \( x \) and \( y \), and then multiply these by the partial derivatives of \( x \) and \( y \) with respect to \( r \).
The same concept is applied to calculate \( \frac{\partial u}{\partial s} \).
This method leverages substitution from \( x \) and \( y \) in terms of \( r \) and \( s \) before performing the differentiation.

The Substitution Method simplifies the function before differentiation. By substituting variables, we transform the original function into a more straightforward expression.
For the given problem, we substitute \( x = 3r - s \) and \( y = r + 2s \) directly into \( u = x^2 - y^2 \). This gives us the new expression in terms of \( r \) and \( s \): \[ u = (3r - s)^2 - (r + 2s)^2 \]
Simplifying this expression:
\[ u = 9r^2 - 6rs + s^2 - (r^2 + 4rs + 4s^2) = 8r^2 - 10rs - 3s^2 \]
Now, with \( u \) written fully in terms of \( r \) and \( s \), differentiating becomes more direct.
This step reduces the error margin and ensures simpler computations in further differentiation steps.

Differentiation is the process of finding the rate at which a function changes with respect to one of its variables. In the context of partial derivatives, it's about how one variable impacts the overall function while keeping others constant.
After substitution, the derivative of \( u = 8r^2 - 10rs - 3s^2 \) with respect to \( r \) is found by treating \( s \) as a constant:
\[ \frac{\partial u}{\partial r} = \frac{\partial}{\partial r} (8r^2 - 10rs - 3s^2) = 16r - 10s \]
Similarly, the partial derivative with respect to \( s \) is:
\[ \frac{\partial u}{\partial s} = \frac{\partial}{\partial s} (8r^2 - 10rs - 3s^2) = -10r - 6s \]
Thus, differentiation combined with prior substitution boils down the problem, making the core principles and results much clearer and more manageable.
Understanding differentiation is crucial, as it forms the backbone of calculus and is widely used in various fields such as physics, engineering, and economics.