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

Explanation

To differentiate the function \( y = \sin^n(ax^2 + bx + c) \), we will use the chain rule and the product rule. Here’s a step-by-step breakdown of the solution:

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

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

### Step 3: Differentiate the inner function
Now we need to find \( \frac{du}{dx} \):
\[ u = ax^2 + bx + c \]
Differentiating \( u \) with respect to \( x \):
\[ \frac{du}{dx} = 2ax + b \]

### Step 4: Substitute back into the derivative
Now we can substitute \( u \) and \( \frac{du}{dx} \) back into our derivative:
\[ \frac{dy}{dx} = n \sin^{n-1}(ax^2 + bx + c) \cdot (2ax + b) \]

### Step 5: Include the cosine term
Since we need to differentiate \( \sin(ax^2 + bx + c) \), we also need to multiply by the derivative of \( \sin(u) \):
\[ \frac{dy}{dx} = n \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot \frac{du}{dx} \]
Thus, we have:
\[ \frac{dy}{dx} = n \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot (2ax + b) \]

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

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