What is the derivative of `f(x)=sqrt(1-x^(2))` with respect to `g(x)=sin^(-1)x,` where `|x|ne1`?

Options

A. `x`
B. `-x`
C. `(x)/(1-x^(2))`
D. `-(x)/(1-x^(2))`

Explanation

To find the derivative of \( f(x) = \sqrt{1 - x^2} \) with respect to \( g(x) = \sin^{-1} x \), we will use the chain rule. The chain rule states that if we have two functions, \( f \) and \( g \), the derivative of \( f \) with respect to \( g \) can be expressed as:

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

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

We start by differentiating \( f(x) \):

\[ f(x) = \sqrt{1 - x^2} \]

Using the chain rule and the power rule:

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

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

Next, we differentiate \( g(x) = \sin^{-1} x \):

\[ \frac{dg}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

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

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

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

This simplifies to:

\[ \frac{df}{dg} = -x \]

### Final Answer

Thus, the derivative of \( f(x) = \sqrt{1 - x^2} \) with respect to \( g(x) = \sin^{-1} x \) is:

\[ \frac{df}{dg} = -x \]