The slope of the tangent of the curve `y=int_0^x (dx)/(1+x^3)` at the point where `x = 1` is

Options

A. ` (1)/(2)`
B. 1
C. `(1)/(4)`
D. non-existent

Explanation

To find the slope of the tangent of the curve defined by the integral \( y = \int_0^x \frac{dt}{1+t^3} \) at the point where \( x = 1 \), we can follow these steps:

### Step 1: Define the function
We start with the function defined by the integral:
\[ y = \int_0^x \frac{dt}{1+t^3} \]

### Step 2: Differentiate using Leibniz's rule
To find the slope of the tangent, we need to differentiate \( y \) with respect to \( x \). According to Leibniz's rule for differentiation under the integral sign:
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \int_0^x \frac{dt}{1+t^3} \right) = \frac{1}{1+x^3} \cdot \frac{d}{dx}(x) - \frac{1}{1+0^3} \cdot \frac{d}{dx}(0) \]
Since \( \frac{d}{dx}(0) = 0 \), the second term vanishes.

### Step 3: Simplify the derivative
Thus, we have:
\[ \frac{dy}{dx} = \frac{1}{1+x^3} \cdot 1 = \frac{1}{1+x^3} \]

### Step 4: Evaluate the derivative at \( x = 1 \)
Now we substitute \( x = 1 \) into the derivative to find the slope of the tangent:
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{1}{1+1^3} = \frac{1}{1+1} = \frac{1}{2} \]

### Conclusion
The slope of the tangent to the curve at the point where \( x = 1 \) is:
\[ \frac{1}{2} \]