Math Problem Statement

Given the piecewise function y = {x*sin(1/x) if x ≠ 0, 0 if x = 0}, find the derivative y'.

Solution

The problem defines a piecewise function for yy, and the task is to find yy', the derivative. The function is as follows:

\begin{cases} x \sin\left(\frac{1}{x}\right), & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$$ ### Steps to Find $$y'$$: #### 1. **Case 1: $$x \neq 0$$** For $$x \neq 0$$, the function is $$y = x \sin\left(\frac{1}{x}\right)$$. Using the product rule: $$y' = \frac{d}{dx}\left(x\right)\sin\left(\frac{1}{x}\right) + x \frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right)$$ - The derivative of $$x$$ is $$1$$. - The derivative of $$\sin\left(\frac{1}{x}\right)$$ is: $$\frac{d}{dx}\left(\sin\left(\frac{1}{x}\right)\right) = \cos\left(\frac{1}{x}\right) \cdot \frac{d}{dx}\left(\frac{1}{x}\right) = -\frac{\cos\left(\frac{1}{x}\right)}{x^2}$$ Substitute back into the product rule formula: $$y' = \sin\left(\frac{1}{x}\right) + x \cdot \left(-\frac{\cos\left(\frac{1}{x}\right)}{x^2}\right)$$ $$y' = \sin\left(\frac{1}{x}\right) - \frac{\cos\left(\frac{1}{x}\right)}{x}$$ Thus, for $$x \neq 0$$: $$y' = \sin\left(\frac{1}{x}\right) - \frac{\cos\left(\frac{1}{x}\right)}{x}.$$ #### 2. **Case 2: $$x = 0$$** For $$x = 0$$, the function is $$y = 0$$, which is a constant. The derivative of a constant is $$0$$: $$y' = 0, \quad \text{if } x = 0.$$ #### 3. **Continuity and Differentiability at $$x = 0$$:** We need to verify if the derivative is continuous at $$x = 0$$. To do this, calculate the left-hand and right-hand limits of $$y'$$ as $$x \to 0$$. Since $$y = x \sin\left(\frac{1}{x}\right)$$ for $$x \neq 0$$, we also need to confirm differentiability at $$x = 0$$. This requires additional checks. - The limit of $$\sin\left(\frac{1}{x}\right)$$ oscillates as $$x \to 0$$, and thus, $$y'$$ does not settle to a single value as $$x \to 0$$. - Therefore, $$y'$$ is not continuous at $$x = 0$$, and the piecewise derivative remains valid as written. ### Final Answer: $$y' = \begin{cases} \sin\left(\frac{1}{x}\right) - \frac{\cos\left(\frac{1}{x}\right)}{x}, & \text{if } x \neq 0, \\ 0, & \text{if } x = 0. \end{cases}$$ --- Would you like me to explain the continuity or differentiability check at $$x = 0$$ in more detail? ### Related Questions: 1. What is the physical significance of the derivative in a piecewise function like this one? 2. How does the product rule apply in general to functions with oscillatory behavior? 3. What are the implications of discontinuity in the derivative for this function? 4. Can this function be made differentiable at $$x = 0$$ by redefining $$y(0)$$? 5. How would the second derivative of this piecewise function be calculated? ### Tip: When dealing with piecewise functions, always check both the value and behavior of derivatives at the boundaries to ensure continuity and differentiability.

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Math Problem Analysis

Mathematical Concepts

Piecewise Functions
Derivatives
Product Rule
Continuity
Differentiability

Formulas

y = x*sin(1/x) for x ≠ 0
Product rule: (uv)' = u'v + uv'
Derivative of sin(f(x)): cos(f(x)) * f'(x)

Theorems

Product Rule of Differentiation

Suitable Grade Level

Undergraduate (Calculus I/II)