The slope of the normal to the curve `y=2x^(2)+3 sin x` at x = 0 is

Options

A. 3
B. `(1)/(3)`
C. `-3`
D. `-(1)/(3)`

Answer

To find the slope of the normal to the curve \( y = 2x^2 + 3 \sin x \) at \( x = 0 \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y \) with respect to \( x \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 + 3 \sin x) \]

Using the rules of differentiation:
- The derivative of \( 2x^2 \) is \( 4x \).
- The derivative of \( 3 \sin x \) is \( 3 \cos x \).

Thus, we have:

\[ \frac{dy}{dx} = 4x + 3 \cos x \]

### Step 2: Evaluate the derivative at \( x = 0 \)
Next, we evaluate the derivative at \( x = 0 \) to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=0} = 4(0) + 3 \cos(0) \]

Since \( \cos(0) = 1 \):

\[ \frac{dy}{dx} \bigg|_{x=0} = 0 + 3(1) = 3 \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, we calculate:

\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{3} \]

### Final Answer
The slope of the normal to the curve \( y = 2x^2 + 3 \sin x \) at \( x = 0 \) is:

\[ \boxed{-\frac{1}{3}} \]