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

Answer

To find the second derivative of the function \( y = \int_{4}^{4x^2} t^4 e^{4t} dt \), we will use the Fundamental Theorem of Calculus and the chain rule.

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

Using the Fundamental Theorem of Calculus, we differentiate the integral with respect to \( x \):

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

By the chain rule, this becomes:

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

Where \( f(t) = t^4 e^{4t} \). Since \( 4 \) is a constant, its derivative is \( 0 \).

Now we compute \( f(4x^2) \):

\[ f(4x^2) = (4x^2)^4 e^{4(4x^2)} = 256 x^8 e^{16x^2} \]

Now, differentiate \( 4x^2 \):

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

Putting it all together, we have:

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

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

Next, we differentiate \( \frac{dy}{dx} \):

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

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 each derivative:

1. \( \frac{d}{dx}(x^9) = 9x^8 \)
2. To differentiate \( e^{16x^2} \), we use the chain rule:

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

Now substituting back into the equation:

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

Factoring out common terms:

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

Finally, we can factor out \( x^8 \):

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

### Final Answer

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