Question:
If \( f(x) = |x| \sin |x| \), then \( f'(x) \) is equal to
If \( f(x) = |x| \sin |x| \), then \( f'(x) \) is equal to
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When differentiating absolute value functions, consider the piecewise nature of the function and apply the product and chain rules accordingly.
Updated On: Apr 22, 2026
- \( \left( \frac{\pi}{4} \right)^{1/2} \sqrt{2} \log \frac{4}{\pi} + 2\sqrt{2} \)
- \( \left( \frac{\pi}{4} \right)^{1/2} \sqrt{2} \log \frac{4}{\pi} \)
- \( \left( \frac{\pi}{4} \right)^{1/2} \sqrt{2} \log \frac{4}{\pi} + 2\sqrt{2} \)
- \( \left( \sqrt{2} \right) \log \frac{\pi}{4} + \frac{2\sqrt{2}}{\pi} \)
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The Correct Option is A
Solution and Explanation
Step 1: Recognize the function and its components.
We are given the function \( f(x) = |x| \sin |x| \). The absolute value function \( |x| \) implies that the function behaves differently for positive and negative values of \( x \). We must differentiate the function piecewise.
Step 2: Differentiate the function for \( x>0 \).
When \( x>0 \), we have: \[ f(x) = x \sin x \] Using the product rule, the derivative is: \[ f'(x) = \sin x + x \cos x \]
Step 3: Differentiate the function for \( x<0 \).
When \( x<0 \), we have: \[ f(x) = -x \sin(-x) = x \sin x \] Again, using the product rule, the derivative is: \[ f'(x) = \sin x + x \cos x \]
Step 4: Apply the absolute value.
The function is piecewise defined, and we need to adjust the derivative for each case. Therefore, for both \( x>0 \) and \( x<0 \), the derivative is the same: \[ f'(x) = \sin x + x \cos x \]
Step 5: Conclusion.
Thus, the derivative of \( f(x) = |x| \sin |x| \) is \( f'(x) = \sin x + x \cos x \), corresponding to option (A).
We are given the function \( f(x) = |x| \sin |x| \). The absolute value function \( |x| \) implies that the function behaves differently for positive and negative values of \( x \). We must differentiate the function piecewise.
Step 2: Differentiate the function for \( x>0 \).
When \( x>0 \), we have: \[ f(x) = x \sin x \] Using the product rule, the derivative is: \[ f'(x) = \sin x + x \cos x \]
Step 3: Differentiate the function for \( x<0 \).
When \( x<0 \), we have: \[ f(x) = -x \sin(-x) = x \sin x \] Again, using the product rule, the derivative is: \[ f'(x) = \sin x + x \cos x \]
Step 4: Apply the absolute value.
The function is piecewise defined, and we need to adjust the derivative for each case. Therefore, for both \( x>0 \) and \( x<0 \), the derivative is the same: \[ f'(x) = \sin x + x \cos x \]
Step 5: Conclusion.
Thus, the derivative of \( f(x) = |x| \sin |x| \) is \( f'(x) = \sin x + x \cos x \), corresponding to option (A).
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