Let f(x)=2cos(x)sin−1(x). Find f′(x).
f′(x)=

To find the derivative of \( f(x) = 2 \cos(x) \sin^{-1}(x) \), we can apply the product rule, as \( f(x) \) is a product of two functions: \( u(x) = 2 \cos(x) \) and \( v(x) = \sin^{-1}(x) \).

### Steps:

1. **Identify \( u(x) \) and \( v(x) \):**
   \[
   u(x) = 2 \cos(x) \quad \text{and} \quad v(x) = \sin^{-1}(x)
   \]

2. **Compute \( u'(x) \):**
   \[
   u(x) = 2 \cos(x) \Rightarrow u'(x) = 2 \cdot (-\sin(x)) = -2 \sin(x)
   \]

3. **Compute \( v'(x) \):**
   Using the derivative of \( \sin^{-1}(x) \):
   \[
   v(x) = \sin^{-1}(x) \Rightarrow v'(x) = \frac{1}{\sqrt{1 - x^2}}
   \]

4. **Apply the product rule:**  
   The product rule states that \( f'(x) = u'(x)v(x) + u(x)v'(x) \):
   \[
   f'(x) = (-2 \sin(x)) \cdot \sin^{-1}(x) + (2 \cos(x)) \cdot \frac{1}{\sqrt{1 - x^2}}
   \]

5. **Combine terms:**
   \[
   f'(x) = -2 \sin(x) \sin^{-1}(x) + \frac{2 \cos(x)}{\sqrt{1 - x^2}}
   \]

### Final Answer:
\[
f'(x) = -2 \sin(x) \sin^{-1}(x) + \frac{2 \cos(x)}{\sqrt{1 - x^2}}
\]

Would you like further details on any step?

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1. How is the product rule applied in differentiation?
2. What are common derivatives of inverse trigonometric functions?
3. Why is the chain rule necessary in derivatives involving trigonometric functions?
4. What is the significance of \( \sin^{-1}(x) \) in trigonometric identities?
5. How does understanding derivative rules simplify complex functions?

**Tip:** Remember that inverse trigonometric functions often involve square roots in their derivatives; be mindful of the domain restrictions for these functions.

Let f(x) = 2cos(x)sin⁻¹(x). Find f'(x).

Find the derivative of f(x) = 2cos(x)sin⁻¹(x) with a step-by-step solution using the product rule. This problem includes differentiation techniques for trigonometric and inverse trigonometric functions, suitable for advanced high school calculus students.

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