91Ó°ÊÓ

Partial derivatives are a fundamental concept in calculus, especially in multivariable functions. They represent how a function changes as one of its variables changes, while keeping the other variables constant.
In our exercise, we have a function of two variables, \(f(x,y) = x \, \sin(xy)\). To find the partial derivatives, we compute the derivative of \(f\) with respect to each variable independently.
For the partial derivative with respect to \(x\), we denote it as \(f_{x}\) or \(\frac{\partial f}{\partial x}\). Similarly, for \(y\), it is denoted as \(f_{y}\) or \(\frac{\partial f}{\partial y}\).

• In the problem, \(f_{x}\) involves applying the product rule, resulting in \(\sin(xy) + xy \cos(xy)\).
• For \(f_{y}\), we use the chain rule, giving us \(x^{2} \cos(xy)\).

Evaluating these at the given points helps in approximating the function value near those points.

The product rule is a formula used to find the derivative of the product of two functions. If we have two functions, \(u(x)\) and \(v(x)\), the product rule states:
The derivative of \(u(x)v(x)\) is \(u'(x)v(x) + u(x)v'(x)\).
In our example, \(f(x,y) = x \sin(xy)\), we consider \(u(x) = x\) and \(v(x) = \sin(xy)\). Applying the product rule, we get:

• The derivative of \(x\) with respect to \(x\) is 1.
• The derivative of \(\sin(xy)\) with respect to \(x\) needs the chain rule and results in \(y \cos(xy)\).

Combining these, we obtain: \( f_{x} = \sin(xy) + xy \cos(xy)\).
This result is key for understanding how the function changes with respect to \(x\).

The chain rule is used for differentiating a composite function. If a function \(z\) depends on \(u\), and \(u\) depends on \(x\), then the chain rule states:
\(\frac{dz}{dx} = \frac{dz}{du} \cdot \frac{du}{dx}\).
In the context of our problem, one of our partial derivatives requires the chain rule. When evaluating \( \frac{\partial}{\partial y} (x\sin(xy))\), we recognize that \(xy\) is a composite function.

• The outer function is \(\sin\), and its derivative is \(\cos\).
• The inner function is \(xy\), and its derivative with respect to \(y\) is \(x\).

Applying the chain rule here, we multiply these derivatives:
Hence, \( f_{y} = x^{2} \cos(xy)\).
This helps understand how the function changes as \(y\) changes.

Linear approximation, also known as linearization, helps estimate the value of a function near a given point using its tangent plane. The formula for linear approximation at a point \((a,b)\) is:
\(L(x, y) = f(a, b) + f_{x}(a, b)(x - a) + f_{y}(a, b)(y - b)\).
In the given exercise, we use this formula to approximate \(z\) near \((1,0)\). The steps are:

• First, compute \(f(a,b)\) at the given points.
• Then find \(f_{x}(a,b)\) and \(f_{y}(a,b)\).
• Substitute all these values into the linearization formula.

This results in \(L(x, y) = y\), which allows us to estimate \(z(0.99, 0.02)\) as 0.02.
Linear approximation simplifies complex functions into linear equations, making them easier to work with near specific points.