If `x= sin^(-1)(3t-4t^(3))` and `y=cos^(-1)(sqrt(1-t^(2)))`, then `(dy)/(dx)` is equal to

Options

A. `1/2`
B. `2/5`
C. `3/2`
D. `1/3`

Answer

To find \(\frac{dy}{dx}\) given \(x = \sin^{-1}(3t - 4t^3)\) and \(y = \cos^{-1}(\sqrt{1 - t^2})\), we will use the chain rule and implicit differentiation.

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

We start with:
\[ x = \sin^{-1}(3t - 4t^3) \]
To differentiate \(x\) with respect to \(t\), we use the derivative of \(\sin^{-1}(u)\), which is \(\frac{1}{\sqrt{1 - u^2}} \cdot \frac{du}{dt}\).

Let \(u = 3t - 4t^3\). Then:
\[ \frac{du}{dt} = 3 - 12t^2 \]

Now, applying the chain rule:
\[ \frac{dx}{dt} = \frac{1}{\sqrt{1 - (3t - 4t^3)^2}} \cdot (3 - 12t^2) \]

### Step 2: Differentiate \(y\) with respect to \(t\)

Next, we differentiate \(y\):
\[ y = \cos^{-1}(\sqrt{1 - t^2}) \]
The derivative of \(\cos^{-1}(v)\) is \(-\frac{1}{\sqrt{1 - v^2}} \cdot \frac{dv}{dt}\).

Let \(v = \sqrt{1 - t^2}\). Then:
\[ \frac{dv}{dt} = \frac{-t}{\sqrt{1 - t^2}} \]

Now, applying the chain rule:
\[ \frac{dy}{dt} = -\frac{1}{\sqrt{1 - (1 - t^2)}} \cdot \left(\frac{-t}{\sqrt{1 - t^2}}\right) = \frac{t}{\sqrt{t^2}} \]
Since \(\sqrt{t^2} = |t|\), we have:
\[ \frac{dy}{dt} = \frac{t}{|t|} = 1 \text{ (for } t > 0\text{)} \]

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

Now we can find \(\frac{dy}{dx}\) using the relationship:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]

Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{1}{\frac{1}{\sqrt{1 - (3t - 4t^3)^2}} \cdot (3 - 12t^2)} \]

This simplifies to:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - (3t - 4t^3)^2}}{3 - 12t^2} \]

### Final Result

Thus, the final answer for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\sqrt{1 - (3t - 4t^3)^2}}{3 - 12t^2} \]