This problem illustrates that the derivative of a differentiable function might not even be continuous. Let Ja² sin(1/x), if x # 0 f(x) = if x = 0. For this problem you may assume as known that sin(x) is differentiable on all of R, sin'(x) | sin(x)| < 1 for all x, cos(x) is continuous at all x, and cos(0) = 1. (These, I believe, are the only facts concerning sin(x) you need to use, but if you think you need other facts for your solution, ask me about them.) (a) Use the Theorem about dervatives of sums, products, etc. and the Chain Rule to prove f is differentiable at all x + 0, and find a fomula for f'(x) that is valid for all x # 0. (b) Use the definition of the derivative to prove f(x) is differentiable at 0 and f'(0) = 0. (c) Parts (a) and (b) show that f'(x) is defined for all x. Prove f'(x) is not continuous at x = 0 by showing that lim,-40 f'(x) = 1. cos(x) for all x,

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
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real analysis

This problem illustrates that the derivative of a differentiable function might not even be continuous.

Let  
\[ 
f(x) = 
\begin{cases} 
x^2 \sin(1/x), & \text{if } x \neq 0 \\ 
0, & \text{if } x = 0.
\end{cases} 
\]

For this problem, you may assume as known that \(\sin(x)\) is differentiable on all of \(\mathbb{R}\), \(\sin'(x) = \cos(x)\) for all \(x\), \(|\sin(x)| \leq 1\) for all \(x\), \(\cos(x)\) is continuous at all \(x\), and \(\cos(0) = 1\). (These, I believe, are the only facts concerning \(\sin(x)\) you need to use, but if you think you need other facts for your solution, ask me about them.)

(a) Use the Theorem about derivatives of sums, products, etc. and the Chain Rule to prove \(f\) is differentiable at all \(x \neq 0\), and find a formula for \(f'(x)\) that is valid for all \(x \neq 0\).

(b) Use the definition of the derivative to prove \(f(x)\) is differentiable at 0 and \(f'(0) = 0\).

(c) Parts (a) and (b) show that \(f'(x)\) is defined for all \(x\). Prove \(f'(x)\) is not continuous at \(x = 0\) by showing that \(\lim_{x \to 0} f'(x) = 1\).
Transcribed Image Text:This problem illustrates that the derivative of a differentiable function might not even be continuous. Let \[ f(x) = \begin{cases} x^2 \sin(1/x), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0. \end{cases} \] For this problem, you may assume as known that \(\sin(x)\) is differentiable on all of \(\mathbb{R}\), \(\sin'(x) = \cos(x)\) for all \(x\), \(|\sin(x)| \leq 1\) for all \(x\), \(\cos(x)\) is continuous at all \(x\), and \(\cos(0) = 1\). (These, I believe, are the only facts concerning \(\sin(x)\) you need to use, but if you think you need other facts for your solution, ask me about them.) (a) Use the Theorem about derivatives of sums, products, etc. and the Chain Rule to prove \(f\) is differentiable at all \(x \neq 0\), and find a formula for \(f'(x)\) that is valid for all \(x \neq 0\). (b) Use the definition of the derivative to prove \(f(x)\) is differentiable at 0 and \(f'(0) = 0\). (c) Parts (a) and (b) show that \(f'(x)\) is defined for all \(x\). Prove \(f'(x)\) is not continuous at \(x = 0\) by showing that \(\lim_{x \to 0} f'(x) = 1\).
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