If `y=sin(ax+b)`, then what is `(d^(2)y)/(dx^(2))` at `x=-(b)/(a)`. where a and b are constants and `a!=0` ?

Options

A. 0
B. `-1`
C. `sin(a-b)`
D. `sin(a+b)`

Answer

To solve the problem, we need to find the second derivative of the function \( y = \sin(ax + b) \) at the point \( x = -\frac{b}{a} \).

### Step-by-Step Solution:

1. **Write down the function:**
\[ y = \sin(ax + b) \]

2. **Differentiate \( y \) with respect to \( x \) to find the first derivative \( \frac{dy}{dx} \):**
Using the chain rule, we differentiate:
\[ \frac{dy}{dx} = \cos(ax + b) \cdot \frac{d}{dx}(ax + b) = \cos(ax + b) \cdot a = a \cos(ax + b) \]

3. **Differentiate \( \frac{dy}{dx} \) to find the second derivative \( \frac{d^2y}{dx^2} \):**
Differentiate \( \frac{dy}{dx} \):
\[ \frac{d^2y}{dx^2} = \frac{d}{dx}(a \cos(ax + b)) = a \cdot (-\sin(ax + b)) \cdot \frac{d}{dx}(ax + b) = -a^2 \sin(ax + b) \]

4. **Evaluate \( \frac{d^2y}{dx^2} \) at \( x = -\frac{b}{a} \):**
Substitute \( x = -\frac{b}{a} \) into the second derivative:
\[ \frac{d^2y}{dx^2} \bigg|_{x = -\frac{b}{a}} = -a^2 \sin\left(a\left(-\frac{b}{a}\right) + b\right) \]
Simplifying the argument of the sine function:
\[ = -a^2 \sin\left(-b + b\right) = -a^2 \sin(0) \]
Since \( \sin(0) = 0 \):
\[ = -a^2 \cdot 0 = 0 \]

### Final Result:
\[ \frac{d^2y}{dx^2} \bigg|_{x = -\frac{b}{a}} = 0 \]