The derivative of `sin^(-1)(3x-4x^(3))` with respect to `sin^(-1)x,` is

Options

A. `3," for "|x|lt1`
B. `3," for "|x|lt(1)/(2)" and "-3" for "(1)/(2)lt|x|lt1`
C. `-3," for "|x|lt12`
D. `-3," for "|x|le(1)/(2)" and "3" for "(1)/(2)lt|x|lt1`

Explanation

To find the derivative of \( \sin^{-1}(3x - 4x^3) \) with respect to \( \sin^{-1}(x) \), we can follow these steps:

### Step 1: Define Variables
Let:
- \( u = \sin^{-1}(3x - 4x^3) \)
- \( v = \sin^{-1}(x) \)

### Step 2: Express \( u \) in terms of \( v \)
We know that \( v = \sin^{-1}(x) \) implies \( x = \sin(v) \). Therefore, we can substitute \( x \) in \( u \):
\[ u = \sin^{-1}(3\sin(v) - 4\sin^3(v)) \]

### Step 3: Differentiate \( u \) with respect to \( v \)
Using the chain rule, we differentiate \( u \):
\[ \frac{du}{dv} = \frac{1}{\sqrt{1 - (3\sin(v) - 4\sin^3(v))^2}} \cdot \frac{d}{dv}(3\sin(v) - 4\sin^3(v)) \]

### Step 4: Differentiate the inner function
Now, we differentiate \( 3\sin(v) - 4\sin^3(v) \):
\[ \frac{d}{dv}(3\sin(v) - 4\sin^3(v)) = 3\cos(v) - 12\sin^2(v)\cos(v) = 3\cos(v)(1 - 4\sin^2(v)) \]

### Step 5: Substitute back into the derivative
Substituting back, we have:
\[ \frac{du}{dv} = \frac{3\cos(v)(1 - 4\sin^2(v))}{\sqrt{1 - (3\sin(v) - 4\sin^3(v))^2}} \]

### Step 6: Find \( \frac{du}{dv} \)
Now we need to evaluate the expression \( 1 - (3\sin(v) - 4\sin^3(v))^2 \):
Let \( y = 3\sin(v) - 4\sin^3(v) \). Then,
\[ y^2 = (3\sin(v) - 4\sin^3(v))^2 \]
Using the identity \( \sin^2(v) + \cos^2(v) = 1 \), we can simplify the expression.

### Step 7: Final expression for \( \frac{du}{dv} \)
After simplification, we can express \( \frac{du}{dv} \) in terms of \( \sin(v) \) and \( \cos(v) \).

### Step 8: Conclusion
The final result will yield:
\[ \frac{du}{dv} = \pm 3 \]
This means that the derivative of \( \sin^{-1}(3x - 4x^3) \) with respect to \( \sin^{-1}(x) \) is \( 3 \) in certain intervals and \( -3 \) in others.