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 power rule. Here is a step-by-step solution:

### Step 1: Identify the outer and inner functions
The function can be expressed as:
- Outer function: \( u^n \) where \( u = \sin(ax^2 + bx + c) \)
- Inner function: \( u = \sin(v) \) where \( v = ax^2 + bx + c \)

### Step 2: Differentiate the outer function
Using the power rule, the derivative of \( u^n \) with respect to \( u \) is:
\[ \frac{du}{dx} = n u^{n-1} \cdot \frac{du}{dx} \]
Substituting \( u = \sin(ax^2 + bx + c) \):
\[ \frac{dy}{du} = n \sin^{n-1}(ax^2 + bx + c) \]

### Step 3: Differentiate the inner function
Now we differentiate the inner function \( v = ax^2 + bx + c \):
\[ \frac{dv}{dx} = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]

### Step 4: Differentiate the sine function
Next, we differentiate \( u = \sin(v) \):
\[ \frac{du}{dv} = \cos(v) \]
Substituting \( v = ax^2 + bx + c \):
\[ \frac{du}{dv} = \cos(ax^2 + bx + c) \]

### Step 5: Apply the chain rule
Now we apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dv} \cdot \frac{dv}{dx} \]
Substituting the derivatives we found:
\[ \frac{dy}{dx} = n \sin^{n-1}(ax^2 + bx + c) \cdot \cos(ax^2 + bx + c) \cdot (2ax + b) \]

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