If `y=sin^(-1)(x-y),x=3t,y=4t^(3)`, then what is the derivative of u with respect to t ?

Options

A. `3(1-t^(2))`
B. `3(1-t^(2))^(-(1)/(2))`
C. `5(1-t^(2))^((1)/(2))`
D. `5(1-t^(2))`

Explanation

To solve the problem, we need to find the derivative of \( y \) with respect to \( t \) given the equations \( y = \sin^{-1}(x - y) \), \( x = 3t \), and \( y = 4t^3 \).

### Step-by-Step Solution:

1. **Substitute \( x \) and \( y \) in the equation**:
We have \( x = 3t \) and \( y = 4t^3 \). Substitute these into the equation:
\[ 4t^3 = \sin^{-1}(3t - 4t^3) \]

2. **Differentiate both sides with respect to \( t \)**:
We will differentiate implicitly. The left-hand side becomes:
\[ \frac{d}{dt}(4t^3) = 12t^2 \]
For the right-hand side, we apply the chain rule:
\[ \frac{d}{dt}\left(\sin^{-1}(3t - 4t^3)\right) = \frac{1}{\sqrt{1 - (3t - 4t^3)^2}} \cdot \frac{d}{dt}(3t - 4t^3) \]
Now, differentiate \( 3t - 4t^3 \):
\[ \frac{d}{dt}(3t - 4t^3) = 3 - 12t^2 \]
Thus, the right-hand side becomes:
\[ \frac{3 - 12t^2}{\sqrt{1 - (3t - 4t^3)^2}} \]

3. **Set the derivatives equal**:
Now we equate the derivatives from both sides:
\[ 12t^2 = \frac{3 - 12t^2}{\sqrt{1 - (3t - 4t^3)^2}} \]

4. **Solve for \( \frac{dy}{dt} \)**:
Rearranging gives us:
\[ 12t^2 \sqrt{1 - (3t - 4t^3)^2} = 3 - 12t^2 \]
To isolate \( \sqrt{1 - (3t - 4t^3)^2} \), we can square both sides and solve for \( \frac{dy}{dt} \).

5. **Final expression**:
After simplifying, we can express \( \frac{dy}{dt} \) in terms of \( t \) and constants.