`sin^(n)(ax^(2)+bx+c)`

Explanation

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \sin^n(ax^2 + bx + c) \), we will use the chain rule and the power rule of differentiation. Here’s a step-by-step solution:

### Step 1: Identify the function
We have:
\[ y = \sin^n(ax^2 + bx + c) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we apply the chain rule. The derivative of \( \sin^n(u) \) where \( u = ax^2 + bx + c \) is given by:
\[ \frac{dy}{dx} = n \sin^{n-1}(u) \cdot \frac{d}{dx}[\sin(u)] \]

### Step 3: Differentiate \( \sin(u) \)
The derivative of \( \sin(u) \) is \( \cos(u) \cdot \frac{du}{dx} \). Therefore, we have:
\[ \frac{dy}{dx} = n \sin^{n-1}(u) \cdot \cos(u) \cdot \frac{du}{dx} \]

### Step 4: Find \( \frac{du}{dx} \)
Now we need to differentiate \( u = ax^2 + bx + c \):
\[ \frac{du}{dx} = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]

### Step 5: Substitute back into the derivative
Now, substituting \( u \) and \( \frac{du}{dx} \) back into the derivative:
\[ \frac{dy}{dx} = n \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot (2ax + b) \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = n \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot (2ax + b) \]