If `y=sin^(-1)x,z=cos^(-1)sqrt(1-x^(2))," then: "(dy)/(dz)=`

Options

A. `(1)/(sqrt(1-x^(2)))`
B. `cos^(-1)x`
C. 1
D. `tan^(-1)((x)/(sqrt1-x^(2)))`

Answer

To solve the problem, we need to find the derivative \(\frac{dy}{dz}\) where \(y = \sin^{-1}(x)\) and \(z = \cos^{-1}(\sqrt{1 - x^2})\).

### Step-by-Step Solution:

1. **Differentiate \(y\) with respect to \(x\)**:
\[ y = \sin^{-1}(x) \]
The derivative of \(y\) with respect to \(x\) is given by:
\[ \frac{dy}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

2. **Differentiate \(z\) with respect to \(x\)**:
\[ z = \cos^{-1}(\sqrt{1 - x^2}) \]
To differentiate \(z\), we apply the chain rule. The derivative of \(\cos^{-1}(u)\) is \(-\frac{1}{\sqrt{1 - u^2}}\), where \(u = \sqrt{1 - x^2}\).

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

Now, substituting \(u\) into the derivative of \(z\):
\[ \frac{dz}{dx} = -\frac{1}{\sqrt{1 - (\sqrt{1 - x^2})^2}} \cdot \frac{du}{dx} \]
Simplifying the term inside the square root:
\[ \sqrt{1 - (\sqrt{1 - x^2})^2} = \sqrt{1 - (1 - x^2)} = \sqrt{x^2} = |x| \]
Thus, we have:
\[ \frac{dz}{dx} = -\frac{1}{|x|} \cdot \left(-\frac{x}{\sqrt{1 - x^2}}\right) = \frac{1}{|x| \sqrt{1 - x^2}} \]

3. **Use the chain rule to find \(\frac{dy}{dz}\)**:
According to the chain rule:
\[ \frac{dy}{dz} = \frac{dy}{dx} \cdot \frac{dx}{dz} = \frac{dy}{dx} \cdot \frac{1}{\frac{dz}{dx}} \]
Substituting the derivatives we found:
\[ \frac{dy}{dz} = \frac{1}{\sqrt{1 - x^2}} \cdot \frac{\sqrt{1 - x^2}}{1} = 1 \]

### Final Answer:
\[ \frac{dy}{dz} = 1 \]