Differentiate the following w.r.t.x. `sin(msin^(-1)x),|x|lt1`

Answer

To differentiate the function \( y = \sin(m \sin^{-1} x) \) with respect to \( x \), we will use the chain rule. Here are the steps to solve the problem:

### Step 1: Identify the outer and inner functions
We have:
- Outer function: \( \sin(u) \) where \( u = m \sin^{-1} x \)
- Inner function: \( u = m \sin^{-1} x \)

### Step 2: Differentiate the outer function
Using the chain rule, the derivative of \( \sin(u) \) with respect to \( u \) is:
\[ \frac{d}{du} \sin(u) = \cos(u) \]
So, we have:
\[ \frac{d}{dx} \sin(m \sin^{-1} x) = \cos(m \sin^{-1} x) \cdot \frac{du}{dx} \]

### Step 3: Differentiate the inner function
Now we need to differentiate the inner function \( u = m \sin^{-1} x \):
\[ \frac{du}{dx} = m \cdot \frac{d}{dx} \sin^{-1} x \]
The derivative of \( \sin^{-1} x \) is:
\[ \frac{d}{dx} \sin^{-1} x = \frac{1}{\sqrt{1 - x^2}} \]
Thus, we have:
\[ \frac{du}{dx} = m \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 4: Combine the results
Now substituting back into our derivative from Step 2:
\[ \frac{dy}{dx} = \cos(m \sin^{-1} x) \cdot \left( m \cdot \frac{1}{\sqrt{1 - x^2}} \right) \]
This simplifies to:
\[ \frac{dy}{dx} = m \cdot \frac{\cos(m \sin^{-1} x)}{\sqrt{1 - x^2}} \]

### Final Answer
Thus, the derivative of \( y = \sin(m \sin^{-1} x) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = m \cdot \frac{\cos(m \sin^{-1} x)}{\sqrt{1 - x^2}} \]

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