If `int _(0) ^(y) cos t ^(2) dt= int _(0) ^(x ^(2)) (sint)/(t ) dt ` then `(dy)/(dx)` is equal to

Options

A. `(2 sin x)/(x cos y)`
B. `(2 sin x ^(2))/(x cos y ^(2))`
C. `(2 sin x ^(2))/(x cos y)`
D. `(2 sin x)/(x cos y)`

Answer

To solve the problem, we need to differentiate both sides of the given equation using Leibniz's rule for differentiation under the integral sign.

Given:
\[ \int_0^{y} \cos(t^2) \, dt = \int_0^{x^2} \frac{\sin(t)}{t} \, dt \]

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

Using Leibniz's rule, we differentiate the left-hand side (LHS) and the right-hand side (RHS).

**Left-hand side:**
\[ \frac{d}{dx} \left( \int_0^{y} \cos(t^2) \, dt \right) = \cos(y^2) \cdot \frac{dy}{dx} \]

**Right-hand side:**
\[ \frac{d}{dx} \left( \int_0^{x^2} \frac{\sin(t)}{t} \, dt \right) = \frac{\sin(x^2)}{x^2} \cdot \frac{d}{dx}(x^2) = \frac{\sin(x^2)}{x^2} \cdot 2x \]

### Step 2: Set the derivatives equal to each other

Now we equate the derivatives obtained from both sides:
\[ \cos(y^2) \cdot \frac{dy}{dx} = \frac{\sin(x^2)}{x^2} \cdot 2x \]

### Step 3: Solve for \(\frac{dy}{dx}\)

Rearranging the equation to solve for \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{2x \sin(x^2)}{x^2 \cos(y^2)} \]

### Final Result

Thus, the expression for \(\frac{dy}{dx}\) is:
\[ \frac{dy}{dx} = \frac{2 \sin(x^2)}{x \cos(y^2)} \]