if `Y=cos^(-1) "" 2x + cot^(-1) "" 5x + sin^(-1)"" 2x + tan^(-1) "" 5x` then

Options

A. `Y'(0) =0`
B. `Y_2 = Y_4`
C. `Y_5 =Y_6`
D. `Y_1 =Y_3`

Answer

To solve the problem, we need to differentiate the expression given and analyze its behavior. Let's break it down step by step.

### Step 1: Define the function
We have:
\[ Y = \cos^{-1}(2x) + \cot^{-1}(5x) + \sin^{-1}(2x) + \tan^{-1}(5x) \]

### Step 2: Simplify the expression
Using the identities:
1. \(\cos^{-1}(x) + \sin^{-1}(x) = \frac{\pi}{2}\)
2. \(\tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2}\)

We can group the terms:
- For \(\cos^{-1}(2x) + \sin^{-1}(2x)\):
\[ \cos^{-1}(2x) + \sin^{-1}(2x) = \frac{\pi}{2} \]

- For \(\tan^{-1}(5x) + \cot^{-1}(5x)\):
\[ \tan^{-1}(5x) + \cot^{-1}(5x) = \frac{\pi}{2} \]

Thus, we can simplify \(Y\):
\[ Y = \frac{\pi}{2} + \frac{\pi}{2} = \pi \]

### Step 3: Differentiate the function
Now, we differentiate \(Y\) with respect to \(x\):
\[ \frac{dY}{dx} = 0 \]
Since \(Y\) is a constant (\(\pi\)), its derivative is zero.

### Step 4: Evaluate the first derivative at \(x = 0\)
\[ Y' = 0 \quad \text{for all } x \]
Thus,
\[ Y'(0) = 0 \]

### Step 5: Find higher-order derivatives
Since the first derivative is zero, the second derivative will also be zero:
\[ Y'' = 0 \]
\[ Y''(0) = 0 \]

Continuing this process, we find that:
\[ Y^{(n)} = 0 \quad \text{for all } n \geq 1 \]

### Conclusion
Thus, we have:
- \(Y' = 0\)
- \(Y'' = 0\)
- \(Y^{(3)} = 0\)
- \(Y^{(4)} = 0\)
- and so on...

### Final Answer
All derivatives of \(Y\) evaluated at \(x = 0\) are zero.