Math Problem Statement
Turunan pertama
Turunan pertama dari f(x) = |x|(x + 1)^2
Solution
To find the first derivative of the given function , we need to take into account the absolute value and apply the product rule since this is a product of two functions. Here's the process step-by-step:
1. Rewrite the function:
We can break the function into two cases based on the absolute value of :
\begin{cases} x(x + 1)^2 & \text{for } x \geq 0 \\ -x(x + 1)^2 & \text{for } x < 0 \end{cases}$$ Now, we compute the derivative for each case separately. ### 2. Case 1: $$x \geq 0$$ Here, $$f(x) = x(x + 1)^2$$. Apply the product rule for derivatives: \[ f'(x) = \frac{d}{dx}[x] \cdot (x + 1)^2 + x \cdot \frac{d}{dx}[(x + 1)^2] \] First term: $$\frac{d}{dx}[x] = 1, \quad \text{so the first part is } (x + 1)^2$$ Second term: $$\frac{d}{dx}[(x + 1)^2] = 2(x + 1)$$ Thus, the derivative is: $$f'(x) = (x + 1)^2 + 2x(x + 1)$$ Simplify: $$f'(x) = (x + 1)[(x + 1) + 2x] = (x + 1)(3x + 1)$$ ### 3. Case 2: $$x < 0$$ In this case, $$f(x) = -x(x + 1)^2$$. Again, apply the product rule: \[ f'(x) = \frac{d}{dx}[-x] \cdot (x + 1)^2 + (-x) \cdot \frac{d}{dx}[(x + 1)^2] \] First term: $$\frac{d}{dx}[-x] = -1, \quad \text{so the first part is } -(x + 1)^2$$ Second term: $$\frac{d}{dx}[(x + 1)^2] = 2(x + 1)$$ Thus, the derivative is: $$f'(x) = -(x + 1)^2 + 2x(x + 1)$$ Simplify: $$f'(x) = -(x + 1)^2 + 2x(x + 1)$$ $$f'(x) = -(x + 1)(x + 1) + 2x(x + 1)$$ $$f'(x) = (x + 1)(2x - (x + 1)) = (x + 1)(x - 1)$$ ### 4. Final Answer: $$f'(x) = \begin{cases} (x + 1)(3x + 1) & \text{for } x \geq 0 \\ (x + 1)(x - 1) & \text{for } x < 0 \end{cases}$$ Would you like more details on any part of this solution or further clarification? Here are five related questions for further exploration: 1. How would you apply the second derivative to this function? 2. What is the interpretation of the derivative at $$x = 0$$, where the absolute value changes? 3. How can we sketch the graph of $$f'(x)$$ based on this derivative? 4. How would the derivative change if $$f(x)$$ involved $$|x + 1|$$ instead of $$|x|$$? 5. Can we analyze the critical points and concavity of $$f(x)$$ from the derivative? **Tip**: Always be cautious with absolute values when differentiating; they often lead to piecewise functions depending on the sign of the variable inside the absolute value.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Calculus
Derivatives
Piecewise Functions
Formulas
Product Rule: (fg)' = f'g + fg'
Chain Rule: (g(h(x)))' = g'(h(x))h'(x)
Theorems
The derivative of an absolute value function must be treated piecewise
The product rule of differentiation
Suitable Grade Level
Grade 12 or early university calculus