If ` y =sin (ax^(2) +bx +c ) then (dy)/(dx) =`

Options

A. ` (2ax +b) sin (ax^(2) +bx +c)`
B. ` (2ax +b) cos (ax^(2) +bx +c)`
C. ` (2ax +2b) cos (ax^(2) +bx +c)`
D. ` (2ax +2b) sin (ax^(2) +bx +c)`

Answer

To find the derivative of the function \( y = \sin(ax^2 + bx + c) \), we will use the chain rule of differentiation. Here are the steps to solve the problem:

### Step 1: Identify the outer and inner functions
In this case, the outer function is \( \sin(u) \) where \( u = ax^2 + bx + c \).

### Step 2: Differentiate the outer function
The derivative of \( \sin(u) \) with respect to \( u \) is \( \cos(u) \). Thus, we have:
\[ \frac{dy}{du} = \cos(ax^2 + bx + c) \]

### Step 3: Differentiate the inner function
Now we need to differentiate the inner function \( u = ax^2 + bx + c \) with respect to \( x \):
\[ \frac{du}{dx} = \frac{d}{dx}(ax^2 + bx + c) = 2ax + b \]
(Note: The derivative of \( c \) is 0 since it is a constant.)

### Step 4: Apply the chain rule
According to the chain rule, we multiply the derivative of the outer function by the derivative of the inner function:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \cos(ax^2 + bx + c) \cdot (2ax + b) \]

### Step 5: Write the final answer
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = (2ax + b) \cos(ax^2 + bx + c) \]

### Summary of the steps:
1. Identify the outer and inner functions.
2. Differentiate the outer function.
3. Differentiate the inner function.
4. Apply the chain rule to combine the results.
5. Write the final answer.