The slope of the normal to the curve `y=2x^(2)+3sin 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: Find the derivative \( \frac{dy}{dx} \)

The first step is to differentiate the function \( y \) with respect to \( x \).

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

Using the power rule and the derivative of sine:

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

### Step 2: Evaluate the derivative at \( x = 0 \)

Next, we need to find the value of the derivative at \( x = 0 \):

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

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. Thus, we calculate the slope of the normal:

\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx} \bigg|_{x=0}} = -\frac{1}{3} \]

### Final Answer

The slope of the normal to the curve at \( x = 0 \) is:

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