Problem 16 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. $$ f(x, y)=\tan \left(x^{2}+y^{2}\right) ;\left(\frac{\pi}{4},-\frac{\pi}{4}\right) $$

Step-by-Step Solution

Answer

The function \( f(x, y) \) is differentiable at \( \left(\frac{\pi}{4}, -\frac{\pi}{4}\right) \).

Step 1: Determine the partial derivatives

To demonstrate that a function is differentiable at a point, we first need to find the partial derivatives of the function with respect to each variable. Starting with the given function \( f(x, y) = \tan(x^2 + y^2) \), we find:The partial derivative with respect to \( x \):\[ f_x(x, y) = \frac{d}{dx}[\tan(x^2 + y^2)] = \sec^2(x^2+y^2) \cdot 2x \]The partial derivative with respect to \( y \):\[ f_y(x, y) = \frac{d}{dy}[\tan(x^2 + y^2)] = \sec^2(x^2+y^2) \cdot 2y \]

Step 2: Compute the partial derivatives at the given point

Now substitute the point \( \left(\frac{\pi}{4}, -\frac{\pi}{4}\right) \) into the expressions for the partial derivatives.\[ f_x\left(\frac{\pi}{4}, -\frac{\pi}{4}\right) = \sec^2\left(\left(\frac{\pi}{4}\right)^2 + \left(-\frac{\pi}{4}\right)^2\right) \cdot 2 \cdot \frac{\pi}{4} \]\[ = \sec^2\left(\frac{\pi^2}{8} + \frac{\pi^2}{8}\right) \cdot \frac{\pi}{2} = \sec^2\left(\frac{\pi^2}{4}\right) \cdot \frac{\pi}{2} \]\[ f_y\left(\frac{\pi}{4}, -\frac{\pi}{4}\right) = \sec^2\left(\frac{\pi^2}{4}\right) \cdot 2 \cdot \left(-\frac{\pi}{4}\right) = -\sec^2\left(\frac{\pi^2}{4}\right) \cdot \frac{\pi}{2} \]

Step 3: Verify continuity of partial derivatives

For \( f(x, y) \) to be differentiable at \( \left(\frac{\pi}{4}, -\frac{\pi}{4}\right) \), the partial derivatives \( f_x \) and \( f_y \) must be continuous around the point. Since \( \sec \) and its derivatives are continuous wherever \( \tan \) is defined, both \( f_x \) and \( f_y \) are continuous around the given point because the input to the tangent function, \( x^2 + y^2 \), is defined and continuous in this region.

Step 4: Conclude differentiability

Since both partial derivatives are continuous around the point \( \left(\frac{\pi}{4}, -\frac{\pi}{4}\right) \) and are defined at the point, we conclude that \( f(x, y) = \tan(x^2 + y^2) \) is differentiable at \( \left(\frac{\pi}{4}, -\frac{\pi}{4}\right) \).

Key Concepts

Differentiability

In multivariable calculus, a function is said to be differentiable at a point if it behaves like a linear function near that point. The notion of differentiability extends the concept of having a derivative in the single-variable case to functions involving several variables. When we say that a function of two variables, such as \(f(x, y)\), is differentiable at a point \((a, b)\), we're essentially saying that around this point, the function can be well-approximated by a plane, which is the linear part.

The key idea is that for the function to be differentiable, not only must the partial derivatives \(\frac{\partial f}{\partial x}\) and \(\frac{\partial f}{\partial y}\) exist at that point, but they must also be locally continuous near that point. This ensures a smooth transition between the values of the derivatives when taking small steps around \((a, b)\).

• For the function to behave nicely, approximation errors should diminish faster than the distance to the point, indicating the dominance of the linear approximation.
• The expression of the derivative forms a linear map, often visualized as a tangent plane in the context of functions of two variables.

Partial Derivatives

Partial derivatives represent the rate of change of a multivariable function with respect to one variable while keeping the others constant. In the context of multivariable calculus, they are crucial because they provide a way to examine the behavior of a function concerning each of its variables separately.

For the function \(f(x, y) = \tan(x^2 + y^2)\), to find the partial derivatives, we apply the chain rule. Calculating \(f_x\) involves deriving \(tan(x^2 + y^2)\) with respect to \(x\), treating \(y\) as a constant. Similarly, \(f_y\) is found by differentiating with respect to \(y\), treating \(x\) as constant:

• The partial derivative with respect to \(x\) is given by \( f_x(x, y) = \sec^2(x^2+y^2) \cdot 2x \).
• The partial derivative with respect to \(y\) is \( f_y(x, y) = \sec^2(x^2+y^2) \cdot 2y \).

Partial derivatives lay the groundwork for further insights, such as determining the direction of steepest ascent or descent and studying the function's critical points.

Continuity of Functions

Continuity is a core concept to grasp when dealing with differentiability in multivariable calculus. A function is continuous at a point if small changes in input values result in small changes in the output value. This quality is what ensures that both the function itself and its partial derivatives are predictable and free from sudden jumps or gaps.

For \(f(x, y)\) to be differentiable at a point, its partial derivatives must not only exist at that point but must also be continuous in a neighborhood surrounding it. In our problem, this refers to ensuring that both \(f_x\) and \(f_y\) are continuous around the point \(\left(\frac{\pi}{4}, -\frac{\pi}{4}\right)\).

• The function \( \sec \) is continuous wherever \( \tan \) is defined, lending continuity to the partial derivatives as long as \( x^2 + y^2 \) doesn’t lead to an undefined input.
• Because \( x^2 + y^2 \) is a smooth, non-negative function, \(f\) is continuous over the entire plane, provided that tangent inputs are valid.

By ensuring these conditions are met, we verify the differentiability of \(f(x, y)\) at the specified point.