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 \( y = \sin(t^2) \) with respect to \( t \), we will follow these steps:

### Step 1: Find the first derivative \( \frac{dy}{dt} \)

Using the chain rule, we differentiate \( y = \sin(t^2) \):

\[ \frac{dy}{dt} = \cos(t^2) \cdot \frac{d(t^2)}{dt} \]

Now, we know that:

\[ \frac{d(t^2)}{dt} = 2t \]

So, substituting this back, we get:

\[ \frac{dy}{dt} = \cos(t^2) \cdot 2t = 2t \cos(t^2) \]

### Step 2: Find the second derivative \( \frac{d^2y}{dt^2} \)

Now we need to differentiate \( \frac{dy}{dt} = 2t \cos(t^2) \) again with respect to \( t \). We will use the product rule here, which states that if \( u = 2t \) and \( v = \cos(t^2) \), then:

\[ \frac{d(uv)}{dt} = u \frac{dv}{dt} + v \frac{du}{dt} \]

Calculating \( \frac{du}{dt} \) and \( \frac{dv}{dt} \):

1. \( \frac{du}{dt} = 2 \)
2. To find \( \frac{dv}{dt} = \frac{d}{dt}(\cos(t^2)) \), we again use the chain rule:

\[ \frac{dv}{dt} = -\sin(t^2) \cdot \frac{d(t^2)}{dt} = -\sin(t^2) \cdot 2t = -2t \sin(t^2) \]

Now substituting back into the product rule:

\[ \frac{d^2y}{dt^2} = (2t)(-2t \sin(t^2)) + (\cos(t^2))(2) \]

This simplifies to:

\[ \frac{d^2y}{dt^2} = -4t^2 \sin(t^2) + 2 \cos(t^2) \]

### Final Result

Thus, the second derivative \( \frac{d^2y}{dt^2} \) is:

\[ \frac{d^2y}{dt^2} = 2 \cos(t^2) - 4t^2 \sin(t^2) \]