If `y=sin(t^2)`, then `(d^2y)/(dt^2)` will be

Options

A. `2tcos(t^2)`
B. `2cos(t^2)-4t^2sin(t^2)`
C. `4t^2sin(t^2)`
D. `2cos(t^2)`

Explanation

To find the second derivative of \( y = \sin(t^2) \), we will follow these steps:

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

Given:
\[ y = \sin(t^2) \]

Using the chain rule, the first derivative \( \frac{dy}{dt} \) is:
\[ \frac{dy}{dt} = \cos(t^2) \cdot \frac{d(t^2)}{dt} \]
Calculating \( \frac{d(t^2)}{dt} \):
\[ \frac{d(t^2)}{dt} = 2t \]
Thus, we have:
\[ \frac{dy}{dt} = \cos(t^2) \cdot 2t = 2t \cos(t^2) \]

### Step 2: Differentiate \( \frac{dy}{dt} \) to find \( \frac{d^2y}{dt^2} \)

Now we need to differentiate \( \frac{dy}{dt} = 2t \cos(t^2) \) using the product rule. The product rule states:
\[ \frac{d(uv)}{dt} = u \frac{dv}{dt} + v \frac{du}{dt} \]
Let:
- \( u = 2t \) and \( v = \cos(t^2) \)

Calculating \( \frac{du}{dt} \) and \( \frac{dv}{dt} \):
\[ \frac{du}{dt} = 2 \]
For \( \frac{dv}{dt} \), we again use the chain rule:
\[ \frac{dv}{dt} = -\sin(t^2) \cdot \frac{d(t^2)}{dt} = -\sin(t^2) \cdot 2t \]

Now applying the product rule:
\[ \frac{d^2y}{dt^2} = u \frac{dv}{dt} + v \frac{du}{dt} \]
Substituting the values:
\[ \frac{d^2y}{dt^2} = (2t)(-\sin(t^2) \cdot 2t) + (\cos(t^2))(2) \]
Simplifying this gives:
\[ \frac{d^2y}{dt^2} = -4t^2 \sin(t^2) + 2 \cos(t^2) \]

### Final Result

Thus, the second derivative is:
\[ \frac{d^2y}{dt^2} = 2 \cos(t^2) - 4t^2 \sin(t^2) \]