The derivation of ` tan^(-1) "" [(2x)//(1-x^2)]w.r.t sin^(-1)"" {2x//(1+x^2)}` Is ……

Options

A. `(2)/(1+x^2)`
B. `(2)/(sqrt(1-x^2))`
C. `1`
D. None

Answer

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

### Step 1: Define the functions
Let
\[ f(x) = \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \]
and
\[ g(x) = \sin^{-1} \left( \frac{2x}{1 + x^2} \right). \]

### Step 2: Differentiate \( f(x) \)
To differentiate \( f(x) \), we can use the substitution \( x = \tan(\theta) \). Then, we have:
\[ f(x) = \tan^{-1} \left( \tan(2\theta) \right) = 2\theta. \]
Since \( \theta = \tan^{-1}(x) \), we can write:
\[ f(x) = 2 \tan^{-1}(x). \]

Now, we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx} [2 \tan^{-1}(x)] = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2}. \]

### Step 3: Differentiate \( g(x) \)
Next, we differentiate \( g(x) \) using the same substitution \( x = \tan(\theta) \):
\[ g(x) = \sin^{-1} \left( \sin(2\theta) \right) = 2\theta. \]
Thus, we have:
\[ g(x) = 2 \tan^{-1}(x). \]

Now, we differentiate \( g(x) \):
\[ g'(x) = \frac{d}{dx} [2 \tan^{-1}(x)] = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2}. \]

### Step 4: Find the derivative \( \frac{df}{dg} \)
Now we can find the derivative \( \frac{df}{dg} \):
\[ \frac{df}{dg} = \frac{f'(x)}{g'(x)} = \frac{\frac{2}{1 + x^2}}{\frac{2}{1 + x^2}} = 1. \]

### Conclusion
Thus, the derivative of \( \tan^{-1} \left( \frac{2x}{1 - x^2} \right) \) with respect to \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \) is:
\[ \frac{df}{dg} = 1. \]