Differentiate the following w.r.t.x. `e^(cos^(-1)(sqrt(1-x^(2)))),|x|lt1`

Answer

To differentiate the function \( y = e^{\cos^{-1}(\sqrt{1 - x^2})} \) with respect to \( x \), we will use the chain rule and implicit differentiation. Let's go through the steps:

### Step 1: Identify the outer and inner functions
The function can be expressed as:
\[ y = e^{u} \]
where
\[ u = \cos^{-1}(\sqrt{1 - x^2}) \]

### Step 2: Differentiate the outer function
Using the chain rule, we differentiate \( y \) with respect to \( u \):
\[ \frac{dy}{du} = e^{u} \]

### Step 3: Differentiate the inner function
Next, we need to differentiate \( u \) with respect to \( x \):
\[ u = \cos^{-1}(\sqrt{1 - x^2}) \]

Using the chain rule again, we have:
\[ \frac{du}{dx} = \frac{d}{dx} \left( \cos^{-1}(v) \right) \]
where
\[ v = \sqrt{1 - x^2} \]

The derivative of \( \cos^{-1}(v) \) is:
\[ \frac{d}{dv} \left( \cos^{-1}(v) \right) = -\frac{1}{\sqrt{1 - v^2}} \]

Now we need to find \( \frac{dv}{dx} \):
\[ v = (1 - x^2)^{1/2} \]
Using the chain rule:
\[ \frac{dv}{dx} = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}} \]

### Step 4: Combine the derivatives
Now we can combine the derivatives using the chain rule:
\[ \frac{du}{dx} = -\frac{1}{\sqrt{1 - v^2}} \cdot \frac{dv}{dx} \]

Since \( v = \sqrt{1 - x^2} \), we can find \( v^2 \):
\[ v^2 = 1 - x^2 \]
Thus,
\[ 1 - v^2 = x^2 \]
and
\[ \sqrt{1 - v^2} = |x| \]
For \( |x| < 1 \), we can simply write:
\[ \sqrt{1 - v^2} = x \]

So,
\[ \frac{du}{dx} = -\frac{1}{|x|} \cdot \left(-\frac{x}{\sqrt{1 - x^2}}\right) = \frac{1}{\sqrt{1 - x^2}} \]

### Step 5: Apply the chain rule
Now we can find \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = e^{u} \cdot \frac{du}{dx} \]
Substituting \( u \) back in:
\[ \frac{dy}{dx} = e^{\cos^{-1}(\sqrt{1 - x^2})} \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Final Answer
Thus, the derivative of the given function is:
\[ \frac{dy}{dx} = \frac{e^{\cos^{-1}(\sqrt{1 - x^2})}}{\sqrt{1 - x^2}} \]