91Ó°ÊÓ

Partial derivatives are a fundamental part of multivariable calculus. They allow us to examine how a function changes with respect to one variable while keeping other variables constant. These derivatives are crucial when dealing with functions of several variables.

In the given problem, the function is represented as \(z = f(x, y)\). At a particular point, the partial derivatives \(f_x\) and \(f_y\) tell us how the function \(f\) behaves as \(x\) and \(y\) change. Specifically, \(f_x(4,4) = 7\) means that at the point (4, 4), for a small change in \(x\), \(z\) changes approximately 7 times that amount, assuming \(y\) stays constant. Similarly, \(f_y(4,4) = 9\) indicates how \(z\) changes with a small change in \(y\), with \(x\) constant.

Understanding partial derivatives is essential for applying the chain rule in multivariable calculus. They provide the rate of change necessary for linking the dependence between \(x\) and \(y\) with the overarching function \(z = f(x, y)\).

Differentiation is the mathematical process of finding the rate at which something changes. In this context, we focus on differentiating expressions involving \(t\), which stands for time or another parameter. In calculus, basic differentiation rules are applied to find the derivatives of more complex expressions.

Consider the expressions:

• \(x(t) = 2e^{3t} + t^2 - t + 2\)
• \(y(t) = 5e^{3t} + 3t - 1\)

By applying the differentiation rules:

• Derivative of \(x(t)\) with respect to \(t\) is \(\frac{dx}{dt} = 6e^{3t} + 2t - 1\)
• Derivative of \(y(t)\) with respect to \(t\) is \(\frac{dy}{dt} = 15e^{3t} + 3\)

These derivatives are essential inputs for using the chain rule, as they provide intermediary rates of change needed to find \(\frac{dz}{dt}\). Differentiation breaks down complicated functions into manageable pieces, paving the way for more complex calculus operations.

Solving calculus problems involves connecting different principles to find a solution. In this exercise, we're tasked with finding \(\frac{dz}{dt}\), the derivative of \(z\) with respect to \(t\). This problem uses the chain rule, a fundamental tool in calculus problem-solving that links different rates of change.

The chain rule for a function \(z = f(x, y)\) states:

• \(\frac{dz}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt}\)

We're given:

• \(f_x(4, 4) = 7\)
• \(f_y(4, 4) = 9\)

And previously calculated:

• \(\frac{dx}{dt} = 6e^{3t} + 2t - 1\)
• \(\frac{dy}{dt} = 15e^{3t} + 3\)

Plug these into the chain rule formula to find \(\frac{dz}{dt}\). Finally, evaluate this derivative at \(t = 0\) to get the solution, \(197\).

By breaking down the problem into smaller components, calculus problem solving becomes more manageable. It requires understanding each part, from differentiation to the application of rules like the chain rule, to derive the correct solution.