If `y = int_(4)^(4x^(2))t^(4)e^(4t)dt`, find `(d^(2)y)/(dx^(2))`

Answer

To solve the problem of finding the second derivative of \( y \) defined as

\[ y = \int_{4}^{4x^2} t^4 e^{4t} \, dt, \]

we will apply the Fundamental Theorem of Calculus and the Chain Rule.

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

Using the Fundamental Theorem of Calculus, we differentiate \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \int_{4}^{4x^2} t^4 e^{4t} \, dt \right). \]

By the theorem, we have:

\[ \frac{dy}{dx} = f(4x^2) \cdot \frac{d(4x^2)}{dx}, \]

where \( f(t) = t^4 e^{4t} \).

Calculating \( \frac{d(4x^2)}{dx} \):

\[ \frac{d(4x^2)}{dx} = 8x. \]

Thus, we get:

\[ \frac{dy}{dx} = f(4x^2) \cdot 8x = (4x^2)^4 e^{4(4x^2)} \cdot 8x. \]

### Step 2: Simplify \( \frac{dy}{dx} \)

Now we simplify \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = 8x (4x^2)^4 e^{16x^2} = 8x \cdot 256x^8 e^{16x^2} = 2048x^9 e^{16x^2}. \]

### Step 3: Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \)

Now we differentiate \( \frac{dy}{dx} \):

\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(2048x^9 e^{16x^2}). \]

Using the product rule:

\[ \frac{d^2y}{dx^2} = 2048 \left( \frac{d}{dx}(x^9) e^{16x^2} + x^9 \frac{d}{dx}(e^{16x^2}) \right). \]

Calculating \( \frac{d}{dx}(x^9) \):

\[ \frac{d}{dx}(x^9) = 9x^8. \]

Calculating \( \frac{d}{dx}(e^{16x^2}) \) using the chain rule:

\[ \frac{d}{dx}(e^{16x^2}) = e^{16x^2} \cdot \frac{d}{dx}(16x^2) = e^{16x^2} \cdot 32x. \]

Putting it all together:

\[ \frac{d^2y}{dx^2} = 2048 \left( 9x^8 e^{16x^2} + x^9 \cdot 32x e^{16x^2} \right). \]

### Step 4: Factor out common terms

Factoring out \( e^{16x^2} \):

\[ \frac{d^2y}{dx^2} = 2048 e^{16x^2} \left( 9x^8 + 32x^{10} \right). \]

### Final Result

Thus, the second derivative is:

\[ \frac{d^2y}{dx^2} = 2048 e^{16x^2} (9x^8 + 32x^{10}). \]