find the derivative dy/dx for y= sin^-1 (sqrt of x)

To find the derivative \(\frac{dy}{dx}\) for \(y = \sin^{-1}(\sqrt{x})\), we can follow these steps:

1. **Start with the function:**
   \[
   y = \sin^{-1}(\sqrt{x})
   \]

2. **Differentiate both sides with respect to \(x\):**

   To differentiate \(y\) with respect to \(x\), apply the chain rule. Let \(u = \sqrt{x}\), then \(y = \sin^{-1}(u)\).

   The derivative of \(\sin^{-1}(u)\) with respect to \(u\) is:
   \[
   \frac{d}{du} \sin^{-1}(u) = \frac{1}{\sqrt{1-u^2}}
   \]

   Now, differentiate \(u = \sqrt{x}\) with respect to \(x\):
   \[
   \frac{du}{dx} = \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}
   \]

3. **Apply the chain rule:**
   \[
   \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sqrt{1-u^2}} \cdot \frac{1}{2\sqrt{x}}
   \]

4. **Substitute \(u = \sqrt{x}\) back into the expression:**
   \[
   \frac{dy}{dx} = \frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{1}{2\sqrt{x}}
   \]

   Simplify the expression:
   \[
   \frac{dy}{dx} = \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}}
   \]

   Combine the fractions:
   \[
   \frac{dy}{dx} = \frac{1}{2\sqrt{x}\sqrt{1-x}}
   \]

So, the derivative \(\frac{dy}{dx}\) is:
\[
\frac{dy}{dx} = \frac{1}{2\sqrt{x(1-x)}}
\]

Would you like more details, or do you have any questions?

Here are 5 related questions for further exploration:

1. What is the derivative of \(y = \cos^{-1}(\sqrt{x})\)?
2. How does the chain rule apply to trigonometric functions in general?
3. What is the derivative of \(y = \sin^{-1}(x^2)\)?
4. How would you find the second derivative of \(y = \sin^{-1}(\sqrt{x})\)?
5. What is the derivative of \(y = \tan^{-1}(\sqrt{x})\)?

**Tip:** When dealing with inverse trigonometric functions, always remember to apply the chain rule when the argument is more complex than a simple variable.

Discover how to find the derivative dy/dx for the function y = sin^-1(sqrt(x)) using the chain rule. This detailed explanation walks through the differentiation process, covering key calculus concepts like inverse trigonometric functions and their derivatives. Ideal for students in Grades 11-12.

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