If `u=sin^(-1)(x-y),x=3t,y=4t^(3)`, then what is the derivative of u with respect to t?
(A) `3(1-t^2)`
(B) `3(1-t^2)^(-1/2)`
(C) `5(1-t^2)^(-1/2)`
(D) `5(1-t^2)`

Options

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

Answer

To find the derivative of \( u \) with respect to \( t \), we start with the given expressions:

1. **Given:**
\[ u = \sin^{-1}(x - y) \]
where \( x = 3t \) and \( y = 4t^3 \).

2. **Substituting \( x \) and \( y \):**
\[ u = \sin^{-1}(3t - 4t^3) \]

3. **Differentiating \( u \) with respect to \( t \):**
We will use the chain rule for differentiation. The derivative of \( \sin^{-1}(z) \) with respect to \( z \) is \( \frac{1}{\sqrt{1 - z^2}} \). Therefore, we need to find \( \frac{du}{dt} \) as follows:
\[ \frac{du}{dt} = \frac{1}{\sqrt{1 - (3t - 4t^3)^2}} \cdot \frac{d}{dt}(3t - 4t^3) \]

4. **Calculating \( \frac{d}{dt}(3t - 4t^3) \):**
\[ \frac{d}{dt}(3t - 4t^3) = 3 - 12t^2 \]

5. **Putting it all together:**
\[ \frac{du}{dt} = \frac{3 - 12t^2}{\sqrt{1 - (3t - 4t^3)^2}} \]

6. **Simplifying the expression:**
We need to simplify \( 1 - (3t - 4t^3)^2 \):
\[ (3t - 4t^3)^2 = 9t^2 - 24t^4 + 16t^6 \]
Thus,
\[ 1 - (3t - 4t^3)^2 = 1 - (9t^2 - 24t^4 + 16t^6) = 1 - 9t^2 + 24t^4 - 16t^6 \]

7. **Final expression for \( \frac{du}{dt} \):**
Therefore, we have:
\[ \frac{du}{dt} = \frac{3 - 12t^2}{\sqrt{1 - 9t^2 + 24t^4 - 16t^6}} \]

8. **Identifying the correct answer:**
After evaluating the expression, we can compare it with the options provided in the question. The correct answer matches with option (B):
\[ \frac{du}{dt} = 3(1 - t^2)^{-1/2} \]