Differentiate w.r.t x
(i)` b tan ^(-1)(x/a+tan^(-1)"(x)/a)`
(ii) `(sin ^(-1)x)^(2)-(cos^(-1)x)^2`

Answer

To differentiate the given expressions with respect to \( x \), we will solve each part step by step.

### Part (i): Differentiate \( y = b \tan^{-1}\left(\frac{x}{a}\right) + \tan^{-1}\left(\frac{x}{a}\right) \)

1. **Identify the function**:
\[ y = b \tan^{-1}\left(\frac{x}{a}\right) + \tan^{-1}\left(\frac{x}{a}\right) \]

2. **Differentiate using the chain rule**:
The derivative of \( \tan^{-1}(u) \) is given by \( \frac{1}{1 + u^2} \cdot \frac{du}{dx} \). Here, \( u = \frac{x}{a} \).
\[ \frac{dy}{dx} = b \cdot \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{1}{a} + \frac{1}{1 + \left(\frac{x}{a}\right)^2} \cdot \frac{1}{a} \]

3. **Simplify the expression**:
\[ \frac{dy}{dx} = \frac{b}{a(1 + \frac{x^2}{a^2})} + \frac{1}{a(1 + \frac{x^2}{a^2})} \]
Combine the terms:
\[ \frac{dy}{dx} = \frac{b + 1}{a(1 + \frac{x^2}{a^2})} \]

4. **Final expression**:
\[ \frac{dy}{dx} = \frac{(b + 1)a^2}{a^2 + x^2} \]

### Part (ii): Differentiate \( y = (\sin^{-1} x)^2 - (\cos^{-1} x)^2 \)

1. **Identify the function**:
\[ y = (\sin^{-1} x)^2 - (\cos^{-1} x)^2 \]

2. **Use the identity**:
We know that \( \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \). Thus:
\[ (\cos^{-1} x)^2 = \left(\frac{\pi}{2} - \sin^{-1} x\right)^2 \]

3. **Expand the expression**:
\[ y = (\sin^{-1} x)^2 - \left(\frac{\pi}{2} - \sin^{-1} x\right)^2 \]
Using the identity \( a^2 - b^2 = (a - b)(a + b) \):
\[ y = \left(\sin^{-1} x - \left(\frac{\pi}{2} - \sin^{-1} x\right)\right)\left(\sin^{-1} x + \left(\frac{\pi}{2} - \sin^{-1} x\right)\right) \]
Simplifying gives:
\[ y = (2\sin^{-1} x - \frac{\pi}{2})\left(\frac{\pi}{2}\right) \]

4. **Differentiate**:
\[ \frac{dy}{dx} = \frac{\pi}{2} \cdot 2 \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{\pi}{\sqrt{1 - x^2}} \]

### Final Answers:
1. For part (i):
\[ \frac{dy}{dx} = \frac{(b + 1)a^2}{a^2 + x^2} \]

2. For part (ii):
\[ \frac{dy}{dx} = \frac{\pi}{\sqrt{1 - x^2}} \]