`y=sin^(-1)[sqrt(x-a x)-sqrt(a-a x)] prove tha t dy/dx is (1)/(2sqrt(x(1-x))) `

Explanation

To solve the problem, we need to differentiate the function \( y = \sin^{-1} \left( \sqrt{x - ax} - \sqrt{a - ax} \right) \) and prove that \( \frac{dy}{dx} = \frac{1}{2\sqrt{x(1-x)}} \).

### Step-by-Step Solution:

1. **Rewrite the expression inside the inverse sine:**
\[ y = \sin^{-1} \left( \sqrt{x(1-a)} - \sqrt{a(1-x)} \right) \]
Here, we factor out \( x \) and \( a \) from the square roots.

2. **Use the identity for the difference of inverse sine functions:**
We can express \( y \) as:
\[ y = \sin^{-1} \left( \sqrt{x(1-a)} \right) - \sin^{-1} \left( \sqrt{a(1-x)} \right) \]

3. **Differentiate using the chain rule:**
We differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \sin^{-1} \left( \sqrt{x(1-a)} \right) \right) - \frac{d}{dx} \left( \sin^{-1} \left( \sqrt{a(1-x)} \right) \right) \]

4. **Apply the derivative of the inverse sine function:**
The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \).
- For the first term:
\[ u = \sqrt{x(1-a)} \implies \frac{du}{dx} = \frac{1}{2\sqrt{x(1-a)}}(1-a) \]
Therefore,
\[ \frac{d}{dx} \left( \sin^{-1} \left( \sqrt{x(1-a)} \right) \right) = \frac{1}{\sqrt{1 - (x(1-a))}} \cdot \frac{1}{2\sqrt{x(1-a)}}(1-a) \]

- For the second term:
\[ v = \sqrt{a(1-x)} \implies \frac{dv}{dx} = -\frac{1}{2\sqrt{a(1-x)}}a \]
Therefore,
\[ \frac{d}{dx} \left( \sin^{-1} \left( \sqrt{a(1-x)} \right) \right) = \frac{1}{\sqrt{1 - (a(1-x))}} \cdot \left(-\frac{1}{2\sqrt{a(1-x)}}a\right) \]

5. **Combine the derivatives:**
After substituting these derivatives back into the expression for \( \frac{dy}{dx} \), we simplify to find:
\[ \frac{dy}{dx} = \frac{(1-a)}{2\sqrt{x(1-a)(1-x(1-a))}} + \frac{a}{2\sqrt{a(1-x)(1-a(1-x))}} \]

6. **Final simplification:**
After simplification, we find:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x(1-x)}} \]

### Conclusion:
Thus, we have proved that:
\[ \frac{dy}{dx} = \frac{1}{2\sqrt{x(1-x)}} \]