If `x=int_(c^(2))^(tan t)tan^(-1)z dz, y= int_(n)^(sqrt(t))(cos(z^(2)))/(z)dx` then `(dy)/(dx)` is equal to : (where c and n are constants) :

Options

A. `(tan t)/(2 t)`
B. `(cos^(2)t)/(t^(2))`
C. `(cos^(3)t)/(2t^(2))`
D. `(tan t^(2))/(2t^(2))`

Answer

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

### Step 1: Define the integrals
We have:
\[ x = \int_{c^2}^{\tan t} \tan^{-1} z \, dz \]
\[ y = \int_{n}^{\sqrt{t}} \frac{\cos(z^2)}{z} \, dz \]

### Step 2: Differentiate \(x\) with respect to \(t\)
Using Leibniz's rule for differentiation under the integral sign, we differentiate \(x\):
\[ \frac{dx}{dt} = \frac{d}{dt} \left( \int_{c^2}^{\tan t} \tan^{-1} z \, dz \right) \]
According to Leibniz's rule:
\[ \frac{dx}{dt} = \tan^{-1}(\tan t) \cdot \frac{d}{dt}(\tan t) - \tan^{-1}(c^2) \cdot \frac{d}{dt}(c^2) \]
Since \(c\) is a constant, \(\frac{d}{dt}(c^2) = 0\):
\[ \frac{dx}{dt} = t \cdot \sec^2 t \]

### Step 3: Differentiate \(y\) with respect to \(t\)
Now, we differentiate \(y\):
\[ \frac{dy}{dt} = \frac{d}{dt} \left( \int_{n}^{\sqrt{t}} \frac{\cos(z^2)}{z} \, dz \right) \]
Using Leibniz's rule again:
\[ \frac{dy}{dt} = \frac{\cos(\sqrt{t}^2)}{\sqrt{t}} \cdot \frac{d}{dt}(\sqrt{t}) - \frac{\cos(n^2)}{n} \cdot \frac{d}{dt}(n) \]
Again, since \(n\) is a constant, \(\frac{d}{dt}(n) = 0\):
\[ \frac{dy}{dt} = \frac{\cos(t)}{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} = \frac{\cos(t)}{2t} \]

### Step 4: Find \(\frac{dy}{dx}\)
Now we can find \(\frac{dy}{dx}\) using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = \frac{\frac{\cos(t)}{2t}}{t \sec^2 t} \]
This simplifies to:
\[ \frac{dy}{dx} = \frac{\cos(t)}{2t} \cdot \frac{1}{t \sec^2 t} = \frac{\cos(t)}{2t^2} \cdot \cos^2(t) \]
Thus, we have:
\[ \frac{dy}{dx} = \frac{\cos^3(t)}{2t^2} \]

### Final Answer
The final result is:
\[ \frac{dy}{dx} = \frac{\cos^3(t)}{2t^2} \]