Derivative of ` tan^(-1)((x)/(sqrt( 1 - x^(2))))` with respect to
` sin^(-1) (3x - 4x^(3)) ` is

Options

A. `(1)/(sqrt( 1- x^(2)))`
B. `(3)/(sqrt(1 -x^(2)))`
C. 3
D. `(1)/(3)`

Explanation

To find the derivative of \( u = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \) with respect to \( v = \sin^{-1}(3x - 4x^3) \), we will use the chain rule. Let's denote:

- \( u = \tan^{-1}\left(\frac{x}{\sqrt{1 - x^2}}\right) \)
- \( v = \sin^{-1}(3x - 4x^3) \)

We need to find \( \frac{du}{dv} \).

### Step 1: Differentiate \( u \) with respect to \( x \)

Using the chain rule, we first find \( \frac{du}{dx} \):

1. Let \( y = \frac{x}{\sqrt{1 - x^2}} \).
2. Then, \( u = \tan^{-1}(y) \).
3. The derivative of \( u \) with respect to \( y \) is:
\[ \frac{du}{dy} = \frac{1}{1 + y^2} \]
4. Now, we differentiate \( y \):
\[ y = \frac{x}{\sqrt{1 - x^2}} \implies \frac{dy}{dx} = \frac{\sqrt{1 - x^2} \cdot 1 - x \cdot \frac{-x}{\sqrt{1 - x^2}}}{1 - x^2} = \frac{\sqrt{1 - x^2} + \frac{x^2}{\sqrt{1 - x^2}}}{1 - x^2} = \frac{1}{(1 - x^2)^{3/2}} \]
5. Thus, using the chain rule:
\[ \frac{du}{dx} = \frac{1}{1 + y^2} \cdot \frac{dy}{dx} \]
Substitute \( y = \frac{x}{\sqrt{1 - x^2}} \):
\[ 1 + y^2 = 1 + \frac{x^2}{1 - x^2} = \frac{1 - x^2 + x^2}{1 - x^2} = \frac{1}{1 - x^2} \]
Therefore:
\[ \frac{du}{dx} = \frac{1}{\frac{1}{1 - x^2}} \cdot \frac{1}{(1 - x^2)^{3/2}} = (1 - x^2)^{1/2} \]

### Step 2: Differentiate \( v \) with respect to \( x \)

Now, we differentiate \( v \):
1. \( v = \sin^{-1}(3x - 4x^3) \)
2. The derivative of \( v \) with respect to \( x \) is:
\[ \frac{dv}{dx} = \frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot (3 - 12x^2) \]

### Step 3: Use the chain rule to find \( \frac{du}{dv} \)

Using the chain rule:
\[ \frac{du}{dv} = \frac{du}{dx} \cdot \frac{dx}{dv} = \frac{du}{dx} \cdot \frac{1}{\frac{dv}{dx}} \]

Substituting the derivatives we found:
\[ \frac{du}{dv} = \frac{(1 - x^2)^{1/2}}{\frac{1}{\sqrt{1 - (3x - 4x^3)^2}} \cdot (3 - 12x^2)} \]

### Final Answer

Thus, the derivative \( \frac{du}{dv} \) is:
\[ \frac{du}{dv} = \frac{(1 - x^2)^{1/2} \cdot \sqrt{1 - (3x - 4x^3)^2}}{3 - 12x^2} \]