Problem 8
Question
2-8. Let \(f: \mathbf{R} \rightarrow \mathbf{R}^{2}\). Prove that \(f\) is differentiable at \(a \in \mathbf{R}\) if and only if \(f^{1}\) and \(f^{2}\) are, and that in this case $$ f^{\prime}(a)=\left(\begin{array}{c} \left(f^{1}\right)^{\prime}(a) \\ \left(f^{2}\right)^{\prime}(a) \end{array}\right) $$
Step-by-Step Solution
Verified Answer
f is differentiable at a if and only if f^1 and f^2 are differentiable at a. In that case, f'(a) is given by the expression shown.
1Step 1: Understand the Problem
Given a function \(f: \mathbf{R} \rightarrow \mathbf{R}^{2}\), we need to prove that \(f\) is differentiable at \(a \in \mathbf{R}\) if and only if \(f^{1}\) and \(f^{2}\) are differentiable at \(a\), and in that case, the derivative \(f^{\prime}(a)\) is given by: \[ f^{\prime}(a)=\left(\begin{array}{c} \left(f^{1}\right)^{\prime}(a) \ \left(f^{2}\right)^{\prime}(a) \end{array}\right) \]
2Step 2: Define Differentiability
A function \(f\) is differentiable at \(a\) if there exists a linear map \(L\) such that: \[ \lim_{h \to 0} \frac{\|f(a+h) - f(a) - L(h)\|}{|h|} = 0 \] The linear map 'L' in this case is the derivative of \(f\) at \(a\), denoted as \(f^{\prime}(a)\).
3Step 3: Consider \(\f^{1}\) and \(\f^{2}\)
Note that \(f\) can be decomposed into its component functions \(f^{1}\) and \(f^{2}\) such that \(f(x) = (f^{1}(x), f^{2}(x))\). We need to examine these component functions individually regarding their differentiability.
4Step 4: Prove 'If' Condition
Assume that \(f\) is differentiable at \(a\). If \(f\) is differentiable, then there exists \(f^{\prime}(a)\) such that the definition of differentiability holds. By considering the first and second components of \(f(a + h) - f(a)\), we have: \[ \lim_{h \to 0} \frac{|f^{1}(a + h) - f^{1}(a) - L_{1}(h)|}{|h|} = 0 \] \[ \lim_{h \to 0} \frac{|f^{2}(a + h) - f^{2}(a) - L_{2}(h)|}{|h|} = 0 \] where \(L_{1}\) and \(L_{2}\) are the components of the linear map \(L\). This implies \(f^{1}\) and \(f^{2}\) are differentiable at \(a\) with \(L_{1} = (f^{1})^{\prime}(a)\) and \(L_{2} = (f^{2})^{\prime}(a)\).
5Step 5: Prove ‘Only If’ Condition
Now assume that \(f^{1}\) and \(f^{2}\) are differentiable at \(a\). By the definition of differentiability, there exist derivatives \((f^{1})^{\prime}(a)\) and \((f^{2})^{\prime}(a)\). Then:\(L(h) = \left(\begin{array}{c} \left(f^{1}\right)^{\prime}(a) h \right) \ \left(f^{2}\right)^{\prime}(a) h \right) \end{array}\right) \) represents a linear map. Substituting this back into the definition of differentiability for \(f\), we get: \[ \lim_{h \to 0} \frac{|f(a + h) - f(a) - L(h)|}{|h|} = 0 \] proving that \(f\) is differentiable at \(a\).
6Step 6: Combine Results
Having proven both the 'if' and 'only if' parts, we can conclude that \(f\) is differentiable at \(a\) if and only if \(f^{1}\) and \(f^{2}\) are differentiable at \(a\). The derivative \(f^{\prime}(a)\) is then: \[ f^{\prime}(a)=\left(\begin{array}{c} \left(f^{1}\right)^{\prime}(a) \ \left(f^{2}\right)^{\prime}(a) \end{array}\right) \]
Key Concepts
Multivariable CalculusCalculus on ManifoldsComponent FunctionsLinear MapLimit Definition
Multivariable Calculus
Multivariable calculus extends the concepts of single-variable calculus to functions that depend on multiple variables. Instead of dealing with functions of one variable, we study functions where inputs and outputs can be vectors.
For instance, given a function \( f: \mathbf{R} \rightarrow \mathbf{R}^{2} \), we see that one scalar input is mapped to a two-dimensional vector output. Calculus tools like differentiation and integration are adapted to handle functions in multiple dimensions, allowing us to understand the behavior of more complex systems.
For instance, given a function \( f: \mathbf{R} \rightarrow \mathbf{R}^{2} \), we see that one scalar input is mapped to a two-dimensional vector output. Calculus tools like differentiation and integration are adapted to handle functions in multiple dimensions, allowing us to understand the behavior of more complex systems.
Calculus on Manifolds
Calculus on manifolds generalizes the principles of multivariable calculus to more abstract spaces called manifolds. Manifolds are spaces that locally resemble Euclidean space but can have different global structures.
For example, consider the surface of a sphere, which is locally flat like a plane but globally curved. Calculus on manifolds helps us analyze functions and forms on these surfaces. In the exercise, we are still in Euclidean space, but these techniques lay the groundwork for more advanced topics like differential geometry.
For example, consider the surface of a sphere, which is locally flat like a plane but globally curved. Calculus on manifolds helps us analyze functions and forms on these surfaces. In the exercise, we are still in Euclidean space, but these techniques lay the groundwork for more advanced topics like differential geometry.
Component Functions
Component functions break down a multivariable function into simpler parts. If we have a function \( f: \mathbf{R} \rightarrow \mathbf{R}^{2} \) given by \( f(x) = (f^1(x), f^2(x)) \), then \( f^1 \) and \( f^2 \) are its component functions. These are just single-variable functions that represent the individual coordinates of the output vector.
Studying component functions, like in the exercise, allows us to handle each part separately before combining them to understand the whole function. This greatly simplifies the problem of proving differentiability for multivariable functions.
Studying component functions, like in the exercise, allows us to handle each part separately before combining them to understand the whole function. This greatly simplifies the problem of proving differentiability for multivariable functions.
Linear Map
A linear map is a fundamental concept in linear algebra, often used in calculus. It is a function that respects vector addition and scalar multiplication. For differentiability in the context of the exercise, the derivative of a function \( f \) at a point \( a \) is represented by a linear map.
This linear map captures how small changes in input near \( a \) translate to changes in output. In simpler terms, if \( f \) is differentiable, the linear map is the best linear approximation to \( f \) at that point. When handling functions mapping to vectors, this translates to dealing with the components separately, as the exercise demonstrates.
This linear map captures how small changes in input near \( a \) translate to changes in output. In simpler terms, if \( f \) is differentiable, the linear map is the best linear approximation to \( f \) at that point. When handling functions mapping to vectors, this translates to dealing with the components separately, as the exercise demonstrates.
Limit Definition
The limit definition is one of the cornerstones of calculus. To say \( f \) is differentiable at a point \( a \), we use the limit definition:
This concept underpins the exercise's proof that the differentiability of \( f \) is tied to the differentiability of its components.
- For a function \( f(x) \) to be differentiable at a, the following must hold: \( \lim_{h \to 0} \frac{\|f(a+h) - f(a) - L(h)\|}{\|h\|} = 0 \)
This concept underpins the exercise's proof that the differentiability of \( f \) is tied to the differentiability of its components.
Other exercises in this chapter
Problem 6
2-6. Let \(f: \mathbf{R}^{2} \rightarrow \mathbf{R}\) be defined by \(f(x, y)=\sqrt{|x y|}\). Show that \(f\) is not differentiable at \((0,0)\).
View solution Problem 7
2-7. Let \(f: \mathbf{R}^{n} \rightarrow \mathbf{R}\) be a function such that \(|f(x)| \leq|x|^{2}\). Show that \(f\) is differentiable at 0 .
View solution Problem 9
2-9. Two functions \(f, g: \mathbf{R} \rightarrow \mathbf{R}\) are equal up to \(\boldsymbol{n}\) th order at \(a\) if $$ \lim _{h \rightarrow 0} \frac{f(a+h)-g
View solution Problem 12
2-12. A function \(f: \mathbf{R}^{n} \times \mathbf{R}^{m} \rightarrow \mathbf{R}^{p}\) is bilinear if for \(x, x_{1}, x_{2} \in \mathbf{R}^{n}\), \(y, y_{1}, y
View solution