Derivative of `sin^(-1)((t)/(sqrt(1+t^(2))))` with respect to `cos^(-1)((1)/(sqrt(1+t^(2)))` is

Options

A. 1
B. `cot" "t`
C. `tan" "t`
D. 0

Explanation

To find the derivative of \( \sin^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \) with respect to \( \cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) \), we can follow these steps:

### Step 1: Define the Functions
Let:
\[ y = \sin^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \]
\[ z = \cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) \]

### Step 2: Relate the Functions to \( \theta \)
To simplify the calculations, we can set \( t = \tan(\theta) \). Thus:
\[ \frac{t}{\sqrt{1+t^2}} = \frac{\tan(\theta)}{\sqrt{1+\tan^2(\theta)}} = \frac{\tan(\theta)}{\sec(\theta)} = \sin(\theta) \]
So, we can rewrite \( y \):
\[ y = \sin^{-1}(\sin(\theta)) = \theta \]

### Step 3: Rewrite \( z \)
Now, we can find \( z \):
\[ z = \cos^{-1}\left(\frac{1}{\sqrt{1+\tan^2(\theta)}}\right) = \cos^{-1}\left(\cos(\theta)\right) = \theta \]

### Step 4: Find the Derivative
Now we need to find \( \frac{dy}{dz} \):
Since both \( y \) and \( z \) are equal to \( \theta \):
\[ \frac{dy}{dz} = \frac{d\theta}{d\theta} = 1 \]

### Conclusion
Thus, the derivative of \( \sin^{-1}\left(\frac{t}{\sqrt{1+t^2}}\right) \) with respect to \( \cos^{-1}\left(\frac{1}{\sqrt{1+t^2}}\right) \) is:
\[ \frac{dy}{dz} = 1 \]