Problem 13 - Calculus for Biology and Medicine

Question

Show that \(f(x, y)\) is differentiable at the indicated point. $$ f(x, y)=\cos (x+y) ;(0,0) $$

Step-by-Step Solution

Answer

The function is differentiable at the point (0,0).

Step 1: Understand Differentiability

A function is differentiable at a point if it can be approximated by a linear function at that point. Mathematically, for a function \( f(x, y) \) to be differentiable at a point \( (a, b) \), there must exist numbers \( f_x(a, b) \) and \( f_y(a, b) \), which represent the partial derivatives such that: \[ f(x, y) = f(a, b) + f_x(a, b)(x-a) + f_y(a, b)(y-b) + o(\sqrt{(x-a)^2 + (y-b)^2}) \] where \( o(h) \) denotes a function that goes to 0 faster than \( h \) as \( h \to 0 \).

Step 2: Compute Partial Derivatives

To check differentiability at \( (0,0) \), first find the partial derivatives. The partial derivative of \( f(x, y) = \cos(x+y) \) with respect to \( x \) is \( f_x(x, y) = -\sin(x+y) \). Similarly, the partial derivative with respect to \( y \) is \( f_y(x, y) = -\sin(x+y) \). Evaluate both at \( (0,0) \): \( f_x(0,0) = -\sin(0) = 0 \) and \( f_y(0,0) = -\sin(0) = 0 \).

Step 3: Check Differentiability Condition

Substitute the values found into the differentiability condition. We have \( f(0, 0) = \cos(0) = 1 \), and need to check if \( f(x, y) = 1 + 0 \cdot x + 0 \cdot y + o(\sqrt{x^2 + y^2}) \) holds. Simplified, \( f(x, y) = 1 + o(\sqrt{x^2 + y^2}) \). Since \( f(x, y) = \cos(x+y) = 1 - \frac{1}{2}(x+y)^2 + o((x+y)^2) \), which simplifies to \( 1 + o(t) \) for small \( t \), this matches the required form.

Step 4: Conclusion

The function \( f(x, y) = \cos(x+y) \) satisfies the differentiability condition at \( (0, 0) \) because the formula \( f(x, y) = 1 + o(\sqrt{x^2 + y^2}) \) is validated by Taylor expansion. Thus, \( f(x, y) \) is differentiable at the point \( (0, 0) \).

Key Concepts

Partial Derivatives

In multivariable calculus, partial derivatives help us understand how a function changes with respect to each variable independently.
When we have functions of two variables like \(f(x, y)\), partial derivatives are similar to taking derivatives but focusing on one variable at a time, treating the other as constant.
- **Partial Derivative with respect to \(x\):** This measures the rate of change of the function as only \(x\) changes, holding \(y\) constant. - For the function \(f(x, y) = \cos(x+y)\), the partial derivative with respect to \(x\) is \(f_x(x, y) = -\sin(x+y)\).
- **Partial Derivative with respect to \(y\):** This measures the rate of change of the function as only \(y\) changes, holding \(x\) constant. - For \(f(x, y) = \cos(x+y)\), the partial derivative with respect to \(y\) is \(f_y(x, y) = -\sin(x+y)\).
Partial derivatives are crucial in finding the slope of the tangent plane at a point, which is an important step towards determining differentiability at that point.

Differentiability

Differentiability in multivariable calculus extends the idea of having a tangent line in one dimension to having a tangent plane in two dimensions.
A function \(f(x, y)\) is said to be differentiable at a point \((a, b)\) if it can be closely approximated by a linear function around that point.
This is crucial because it lets us approximate small changes in \(f\) using a linear function, just like how derivatives work in single-variable calculus.
- For \(f(x, y) = \cos(x+y)\) at \((0,0)\), differentiability means we need to verify that: \[ f(x, y) = f(0, 0) + f_x(0, 0)(x-0) + f_y(0, 0)(y-0) + o(\sqrt{x^2 + y^2}) \] We computed the partial derivatives:- \(f_x(0,0) = 0\) - \(f_y(0,0) = 0\)Substituting these, we have the simple form \(f(x, y) = 1 + o(\sqrt{x^2 + y^2})\).
The fact that \(f(x, y)\) can be represented this way confirms its differentiability at \((0,0)\).
Differentiability ensures the existence of a linear approximation at a small neighborhood around \((0, 0)\).

Taylor Expansion

Taylor expansion is a method used to approximate functions using polynomials centered around a point.
In multivariable calculus, it plays a key role in understanding differentiability.
The idea is to express \(f(x, y)\) as a sum of terms that include powers of \((x-a)\) and \((y-b)\), where each term's coefficient involves partial derivatives of \(f\) evaluated at the point \((a, b)\).
- For \(f(x, y) = \cos(x+y)\) around the point \((0, 0)\): - The basic Taylor expansion is: - First term: \(f(0, 0) = \cos(0) = 1\).
- Second term: \(-\frac{1}{2}(x+y)^2\), which is derived from the higher order derivatives.
Putting these together, for small inputs \((x, y)\), we get:- \(f(x, y) \approx 1 - \frac{1}{2}(x+y)^2 + o((x+y)^2)\)
This expression corroborates our finding of differentiability, showing the function can be approximated by its Taylor polynomial within higher-order error terms that vanish faster than \(\sqrt{x^2 + y^2}\).
Taylor expansion thus provides a framework for assessing how well a function can be simulated linearly near a point, confirming the differentiability of \(f\) at \((0, 0)\).