Find the slope of the tangent to curve `y=x^3-x+1` at the point whose x-coordinate is 2.

Explanation

To find the slope of the tangent to the curve \( y = x^3 - x + 1 \) at the point where the x-coordinate is 2, we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = x^3 - x + 1 \) with respect to \( x \).

\[ \frac{dy}{dx} = \frac{d}{dx}(x^3) - \frac{d}{dx}(x) + \frac{d}{dx}(1) \]
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