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 definitions of \(x\) and \(y\):

1. **Define \(x\) and \(y\)**:
\[ x = \int_{c^2}^{\tan t} \tan^{-1} z \, dz \]
\[ y = \int_{n}^{\sqrt{t}} \frac{\cos(z^2)}{z} \, dz \]

2. **Differentiate \(x\) with respect to \(t\)**:
Using the Fundamental Theorem of Calculus and the chain rule:
\[ \frac{dx}{dt} = \frac{d}{dt} \left( \int_{c^2}^{\tan t} \tan^{-1} z \, dz \right) = \tan^{-1}(\tan t) \cdot \frac{d}{dt}(\tan t) = \tan^{-1}(\tan t) \cdot \sec^2 t \]
Since \(\tan^{-1}(\tan t) = t\) for \(t\) in the principal range:
\[ \frac{dx}{dt} = t \cdot \sec^2 t \]

3. **Differentiate \(y\) with respect to \(t\)**:
Again, using the Fundamental Theorem of Calculus:
\[ \frac{dy}{dt} = \frac{d}{dt} \left( \int_{n}^{\sqrt{t}} \frac{\cos(z^2)}{z} \, dz \right) = \frac{\cos((\sqrt{t})^2)}{\sqrt{t}} \cdot \frac{d}{dt}(\sqrt{t}) = \frac{\cos(t)}{\sqrt{t}} \cdot \frac{1}{2\sqrt{t}} = \frac{\cos(t)}{2t} \]

4. **Find \(\frac{dy}{dx}\)**:
Using the chain rule:
\[ \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{\frac{\cos(t)}{2t}}{t \cdot \sec^2 t} \]
Simplifying this expression:
\[ \frac{dy}{dx} = \frac{\cos(t)}{2t} \cdot \frac{1}{t \cdot \sec^2 t} = \frac{\cos(t)}{2t^2} \cdot \cos^2 t = \frac{\cos^3 t}{2t^2} \]

5. **Final Result**:
Therefore, the value of \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{\cos^3 t}{2t^2} \]