Derivative of `cos^(-1)sqrt(x)w.r.t.sqrt(1-x)` is

Options

A. `sqrt(x)`
B. `-sqrt(x)`
C. `(1)/(sqrt(x))`
D. `-(1)/(sqrt(x))`

Answer

To find the derivative of \( \cos^{-1}(\sqrt{x}) \) with respect to \( \sqrt{1-x} \), we can use the chain rule. Let's denote:

- \( f(x) = \cos^{-1}(\sqrt{x}) \)
- \( g(x) = \sqrt{1-x} \)

We need to find \( \frac{df}{dg} \).

### Step 1: Find \( \frac{df}{dx} \)

Using the chain rule, the derivative of \( f(x) = \cos^{-1}(\sqrt{x}) \) is given by:

\[ \frac{df}{dx} = -\frac{1}{\sqrt{1 - (\sqrt{x})^2}} \cdot \frac{d}{dx}(\sqrt{x}) \]

Now, we calculate \( \frac{d}{dx}(\sqrt{x}) \):

\[ \frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}} \]

Substituting this back into the derivative of \( f \):

\[ \frac{df}{dx} = -\frac{1}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}} = -\frac{1}{2\sqrt{x(1-x)}} \]

### Step 2: Find \( \frac{dg}{dx} \)

Next, we find the derivative of \( g(x) = \sqrt{1-x} \):

\[ \frac{dg}{dx} = \frac{d}{dx}((1-x)^{1/2}) = \frac{1}{2}(1-x)^{-1/2} \cdot (-1) = -\frac{1}{2\sqrt{1-x}} \]

### Step 3: Find \( \frac{df}{dg} \)

Now, we can find \( \frac{df}{dg} \) using the formula:

\[ \frac{df}{dg} = \frac{df/dx}{dg/dx} \]

Substituting the derivatives we found:

\[ \frac{df}{dg} = \frac{-\frac{1}{2\sqrt{x(1-x)}}}{-\frac{1}{2\sqrt{1-x}}} \]

This simplifies to:

\[ \frac{df}{dg} = \frac{1}{\sqrt{x(1-x)}} \cdot \sqrt{1-x} = \frac{1}{\sqrt{x}} \]

### Final Answer

Thus, the derivative of \( \cos^{-1}(\sqrt{x}) \) with respect to \( \sqrt{1-x} \) is:

\[ \frac{df}{dg} = \frac{1}{\sqrt{x}} \]