The slope of the tangent to the curve `y=x^(3)+x+54` which also passes through the originis

Answer

To find the slope of the tangent to the curve \( y = x^3 + x + 54 \) that passes through the origin, we can follow these steps:

### Step 1: Differentiate the function
We start by finding the derivative of the curve, which gives us the slope of the tangent line at any point \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x^3 + x + 54) \]

### Step 2: Apply the power rule
Using the power rule of differentiation, we differentiate each term:

\[ \frac{dy}{dx} = 3x^2 + 1 + 0 = 3x^2 + 1 \]

### Step 3: Set up the equation for the tangent line
The slope of the tangent line at a point \( (x_0, y_0) \) on the curve is given by \( m = 3x_0^2 + 1 \). Since the tangent line passes through the origin (0,0), we can express the equation of the tangent line using point-slope form:

\[ y - y_0 = m(x - x_0) \]

### Step 4: Substitute \( y_0 \) and \( m \)
From the curve equation, we know:

\[ y_0 = x_0^3 + x_0 + 54 \]

Substituting \( y_0 \) and \( m \):

\[ y - (x_0^3 + x_0 + 54) = (3x_0^2 + 1)(x - x_0) \]

### Step 5: Set \( y = 0 \) and \( x = 0 \)
Since the tangent line passes through the origin, we set \( y = 0 \) and \( x = 0 \):

\[ 0 - (x_0^3 + x_0 + 54) = (3x_0^2 + 1)(0 - x_0) \]

### Step 6: Simplify the equation
This simplifies to:

\[ -(x_0^3 + x_0 + 54) = -(3x_0^3 + x_0) \]

### Step 7: Rearrange the equation
Rearranging gives:

\[ x_0^3 + x_0 + 54 = 3x_0^3 + x_0 \]

### Step 8: Combine like terms
Combining like terms results in:

\[ 54 = 2x_0^3 \]

### Step 9: Solve for \( x_0 \)
Dividing both sides by 2:

\[ x_0^3 = 27 \]

Taking the cube root:

\[ x_0 = 3 \]

### Step 10: Find the slope at \( x_0 = 3 \)
Now we can find the slope \( m \) at \( x_0 = 3 \):

\[ m = 3(3^2) + 1 = 3(9) + 1 = 27 + 1 = 28 \]

### Final Answer
The slope of the tangent to the curve that passes through the origin is:

\[ \boxed{28} \]