91Ó°ÊÓ

The chain rule is a fundamental concept in calculus used when a function is composed of other functions. Essentially, it helps us find the derivative of composite functions. In the exercise, we are dealing with the function inside the logarithm, specifically \( u = \,\ln(1 + x y^{-3/5}) \). Here's how the chain rule works in this context:
- Identify the outer function, which is \( \,\ln(u) \).- Differentiate the outer function: \( \,\frac{d}{du}(\,\ln(u)) = \,\frac{1}{u} \).- Find the inner function derivative: if \( u = 1 + x y^{-3/5} \), then \( \,\frac{du}{dx} = y^{-3/5} \).
By applying the chain rule, we use the derivatives of both the outer and inner functions to find the derivative of the original function with respect to \( x \) or \( y \). This is why for a function like \( \,\ln(1 + x y^{-3/5}) \), the step-by-step differentiation process involves multiplication of these derivatives. It stems from the idea that the rate of change flows through each layer of the functions involved.

The product rule is essential when differentiating functions that are products of two or more functions. In this exercise, our function \( z = x^3 \,\ln(1 + x y^{-3/5}) \) can be seen as a product of two separate functions: \( x^3 \) and \( \,\ln(1 + x y^{-3/5}) \). The product rule states that when you have two functions multiplied together, their derivative is:
\[ \frac{d}{dx}(f(x)g(x)) = f'(x)g(x) + f(x)g'(x) \].
Applying this to our problem:

• Differentiate \( x^3 \) to get \( 3x^2 \).
• Differentiate \( \,\ln(1 + x y^{-3/5}) \) as described in the chain rule section.

These parts combine as per the product rule to give the partial derivative of \( z \) with respect to \( x \). This method allows us to handle the multiplication of functions efficiently without having to expand them algebraically before differentiating.

When dealing with partial derivatives, as in this exercise, we focus on differentiating with respect to one variable at a time. Imagine that when we differentiate with respect to \( x \), \( y \) is treated as a constant, and vice versa. This idea simplifies the differentiation process, making it more manageable. For example:
- To find \( \frac{\partial z}{\partial x} \), hold \( y \) constant, and apply both the chain and product rules to differentiate. - To find \( \frac{\partial z}{\partial y} \), treat \( x \) as a constant, and differentiate accordingly.
Differentiation with respect to a variable in a multivariable function allows us to analyze how that specific variable influences the function independently of others. This method is central to fields like engineering and physics, where understanding the effect of one variable separately is crucial for modeling and predictions. It helps break down complex interactions into simpler, more understandable pieces.