Math Problem Statement
Find the derivative y' of the piecewise function y = {x^2 e^x for x <= 0, sin(x) for x > 0}. Determine if the derivative is continuous at x = 0.
Solution
The given function is a piecewise function defined as:
\begin{cases} x^2 e^x & \text{if } x \leq 0 \\ \sin x & \text{if } x > 0 \end{cases}$$ We need to find $$y'$$, the derivative of $$y$$. Let’s differentiate it piecewise. --- ### 1. **Case 1: $$x \leq 0$$** The function is $$y = x^2 e^x$$. We use the product rule for differentiation: $$y' = \frac{d}{dx}(x^2) \cdot e^x + x^2 \cdot \frac{d}{dx}(e^x)$$ $$y' = 2x e^x + x^2 e^x$$ So, for $$x \leq 0$$: $$y' = e^x (2x + x^2)$$ --- ### 2. **Case 2: $$x > 0$$** The function is $$y = \sin x$$. The derivative of $$\sin x$$ is: $$y' = \cos x$$ --- ### 3. **Behavior at $$x = 0$$:** To ensure the derivative is well-defined at $$x = 0$$, the left-hand and right-hand derivatives must be equal. We compute: #### Left-hand derivative ($$x \to 0^-$$): For $$x \leq 0$$, the derivative is $$y' = e^x (2x + x^2)$$. Substituting $$x = 0$$: $$y' = e^0 (2(0) + 0^2) = 0$$ #### Right-hand derivative ($$x \to 0^+$$): For $$x > 0$$, the derivative is $$y' = \cos x$$. Substituting $$x = 0$$: $$y' = \cos(0) = 1$$ The left-hand derivative ($$0$$) and right-hand derivative ($$1$$) are not equal. Therefore, $$y'$$ is not continuous at $$x = 0$$. --- ### Final Answer: $$y' = \begin{cases} e^x (2x + x^2) & \text{if } x < 0 \\ 1 & \text{if } x = 0 \\ \cos x & \text{if } x > 0 \end{cases}$$ --- Would you like a detailed explanation of any step or further clarification? Here are some follow-up questions: 1. How do you determine whether a piecewise function is differentiable at a point? 2. What are the conditions for continuity and differentiability of piecewise functions? 3. Can we interpret the discontinuity of the derivative at $$x = 0$$ graphically? 4. How does the product rule work in differentiation? 5. How does the behavior of $$e^x$$ affect the derivative for negative $$x$$? **Tip:** Always check for continuity of both the function and its derivative at the boundaries of piecewise functions to confirm differentiability.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Piecewise Functions
Differentiation
Continuity
Limits
Formulas
Product Rule: (uv)' = u'v + uv'
Derivative of sin(x): d/dx[sin(x)] = cos(x)
Derivative of e^x: d/dx[e^x] = e^x
Theorems
Limit Definition of Derivative
Continuity and Differentiability at a Point
Suitable Grade Level
Grade 11-12 or Introductory Calculus Course
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