If `x=int_c^(sint)sin^(-1)z dz ,y =int_k^(sqrt(t))(sinz^2)/zdz,` then `(dy)/(dx)` is equal to (a) `(tant)/(2t)` (b) `(tant)/(t^2)` (c) `t/(2t^2)` (d) `(tan t)/(2t^2)`

Options

A. `(tant)/(2t)`
B. `(tant)/(t^(2))`
C. `(tan t)/(2t^(2))`
D. `(tan t^(2))/(2t^(2))`

Explanation

To solve the problem, we need to find \(\frac{dy}{dx}\) given the definitions of \(x\) and \(y\) in terms of integrals. Let's break down the solution step by step.

### Step 1: Define the integrals
We have:
\[ x = \int_c^{\sin t} \sin^{-1}(z) \, dz \]
\[ y = \int_k^{\sqrt{t}} \frac{\sin(z^2)}{z} \, dz \]

### Step 2: Differentiate \(x\) with respect to \(t\)
Using the Leibniz rule for differentiation under the integral sign, we differentiate \(x\):
\[ \frac{dx}{dt} = \sin^{-1}(\sin t) \cdot \frac{d}{dt}(\sin t) - \sin^{-1}(c) \cdot 0 \]
Since \(\frac{d}{dt}(\sin t) = \cos t\), we have:
\[ \frac{dx}{dt} = \sin^{-1}(\sin t) \cdot \cos t = t \cdot \cos t \]

### Step 3: Differentiate \(y\) with respect to \(t\)
Again, applying the Leibniz rule for \(y\):
\[ \frac{dy}{dt} = \frac{\sin((\sqrt{t})^2)}{\sqrt{t}} \cdot \frac{d}{dt}(\sqrt{t}) - \frac{\sin(k^2)}{k} \cdot 0 \]
Since \(\frac{d}{dt}(\sqrt{t}) = \frac{1}{2\sqrt{t}}\), we have:
\[ \frac{dy}{dt} = \frac{\sin(t)}{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} = \frac{\sin(t)}{2t} \]

### Step 4: Find \(\frac{dy}{dx}\)
Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} \]
We already have \(\frac{dy}{dt} = \frac{\sin(t)}{2t}\) and we need \(\frac{dt}{dx}\):
\[ \frac{dt}{dx} = \frac{1}{\frac{dx}{dt}} = \frac{1}{t \cos t} \]
Thus,
\[ \frac{dy}{dx} = \frac{\sin(t)}{2t} \cdot \frac{1}{t \cos t} = \frac{\sin(t)}{2t^2 \cos t} \]

### Step 5: Simplify \(\frac{dy}{dx}\)
Using the identity \(\tan(t) = \frac{\sin(t)}{\cos(t)}\), we can rewrite:
\[ \frac{dy}{dx} = \frac{\tan(t)}{2t^2} \]

### Conclusion
Thus, the final result is:
\[ \frac{dy}{dx} = \frac{\tan(t)}{2t^2} \]

### Final Answer
The correct option is (d) \(\frac{\tan t}{2t^2}\).