If `y=sin(m sin^(-1)x)`. then what is the value of `d^(2)y//dx^(2)` at `x=0` ?

Options

A. A) m
B. B) `m^(2)`
C. C) `m^(2)+2`
D. D) 0

Answer

To find the value of \( \frac{d^2y}{dx^2} \) at \( x = 0 \) for the function \( y = \sin(m \sin^{-1}(x)) \), we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dx} \)

Given:
\[ y = \sin(m \sin^{-1}(x)) \]

Using the chain rule, the derivative of \( \sin(u) \) is \( \cos(u) \cdot \frac{du}{dx} \), where \( u = m \sin^{-1}(x) \).

First, we compute \( \frac{du}{dx} \):
\[ u = m \sin^{-1}(x) \implies \frac{du}{dx} = m \cdot \frac{1}{\sqrt{1 - x^2}} \]

Now, applying the chain rule:
\[ \frac{dy}{dx} = \cos(m \sin^{-1}(x)) \cdot \frac{du}{dx} = \cos(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}} \]

### Step 2: Find the second derivative \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{m \cos(m \sin^{-1}(x))}{\sqrt{1 - x^2}} \]

This is a product of two functions, so we will use the product rule:
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}\left(\frac{m \cos(m \sin^{-1}(x))}{\sqrt{1 - x^2}}\right) \]

Let \( v = \cos(m \sin^{-1}(x)) \) and \( w = \frac{1}{\sqrt{1 - x^2}} \).

Using the product rule:
\[ \frac{d^2y}{dx^2} = v' w + v w' \]

1. **Finding \( v' \)**:
\[ v' = -\sin(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}} \]

2. **Finding \( w' \)**:
\[ w = (1 - x^2)^{-1/2} \implies w' = \frac{1}{2}(1 - x^2)^{-3/2} \cdot (-2x) = \frac{-x}{(1 - x^2)^{3/2}} \]

Putting it all together:
\[ \frac{d^2y}{dx^2} = \left(-\sin(m \sin^{-1}(x)) \cdot \frac{m}{\sqrt{1 - x^2}}\right) \cdot \frac{1}{\sqrt{1 - x^2}} + \cos(m \sin^{-1}(x)) \cdot \left(\frac{-x}{(1 - x^2)^{3/2}}\right) \]

### Step 3: Evaluate \( \frac{d^2y}{dx^2} \) at \( x = 0 \)

Substituting \( x = 0 \):
- \( \sin^{-1}(0) = 0 \)
- \( \sin(m \sin^{-1}(0)) = \sin(0) = 0 \)
- \( \cos(m \sin^{-1}(0)) = \cos(0) = 1 \)

Thus:
\[ \frac{d^2y}{dx^2} = \left(-0 \cdot \frac{m}{\sqrt{1 - 0^2}}\right) \cdot \frac{1}{\sqrt{1 - 0^2}} + 1 \cdot \left(\frac{-0}{(1 - 0^2)^{3/2}}\right) = 0 + 0 = 0 \]

### Final Answer
\[ \frac{d^2y}{dx^2} \text{ at } x = 0 \text{ is } 0. \]

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