Find the equation of the normal to the curve `y=2x^(3)+3 sin x" at "x=0`.

Answer

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

### Step 1: Find the derivative \( \frac{dy}{dx} \)
Given the function:
\[ y = 2x^3 + 3 \sin x \]
We need to differentiate this with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x^3) + \frac{d}{dx}(3 \sin x) \]
Calculating the derivatives:
\[ \frac{dy}{dx} = 6x^2 + 3 \cos x \]

### Step 2: Evaluate \( \frac{dy}{dx} \) at \( x = 0 \)
Now, we substitute \( x = 0 \) into the derivative:
\[ \frac{dy}{dx} \bigg|_{x=0} = 6(0)^2 + 3 \cos(0) = 0 + 3 = 3 \]

### Step 3: Find the slope of the normal
The slope of the normal line \( m_2 \) is the negative reciprocal of the slope of the tangent line \( m_1 \):
\[ m_2 = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{3} \]

### Step 4: Find the point on the curve at \( x = 0 \)
Next, we need to find the coordinates of the point on the curve at \( x = 0 \):
\[ y = 2(0)^3 + 3 \sin(0) = 0 + 0 = 0 \]
Thus, the point is \( (0, 0) \).

### Step 5: Use the point-slope form to find the equation of the normal
The equation of the normal line can be expressed in point-slope form:
\[ y - y_1 = m_2(x - x_1) \]
Substituting \( (x_1, y_1) = (0, 0) \) and \( m_2 = -\frac{1}{3} \):
\[ y - 0 = -\frac{1}{3}(x - 0) \]
This simplifies to:
\[ y = -\frac{1}{3}x \]

### Step 6: Rearranging the equation
To express this in a more standard form, we can multiply through by 3:
\[ 3y = -x \quad \text{or} \quad x + 3y = 0 \]

### Final Answer
The equation of the normal to the curve at \( x = 0 \) is:
\[ x + 3y = 0 \]