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We are given that the curve is \( y = x^3 \), and the tangent touches the curve at the point \( (1, 1) \) with a gradient (slope) of 3.

### Steps to solve:

1. **Find the derivative of the curve equation:**
   The slope of the tangent to the curve is given by the derivative \( \frac{dy}{dx} \). 
   \[
   y = x^3 \implies \frac{dy}{dx} = 3x^2
   \]
   
2. **Evaluate the slope at the point \( (1, 1) \):**
   We substitute \( x = 1 \) into \( \frac{dy}{dx} = 3x^2 \) to find the slope of the curve at that point.
   \[
   \frac{dy}{dx} \Bigg|_{x=1} = 3(1)^2 = 3
   \]
   So the slope of the tangent at \( (1, 1) \) is indeed 3, which matches the given gradient of the tangent.

3. **Use the point-slope form of the tangent line equation:**
   The point-slope form of the equation of a line is given by:
   \[
   y - y_1 = m(x - x_1)
   \]
   where \( m \) is the slope and \( (x_1, y_1) \) is the point of tangency. Here, \( m = 3 \) and \( (x_1, y_1) = (1, 1) \).
   Substituting these values into the point-slope form:
   \[
   y - 1 = 3(x - 1)
   \]
   
4. **Simplify the equation:**
   \[
   y - 1 = 3x - 3
   \]
   \[
   y = 3x - 2
   \]

Thus, the equation of the tangent is:
\[
y = 3x - 2
\]

Would you like further details or have any questions?

Here are five related questions:
1. How do you compute the derivative of more complex functions?
2. What happens if the point of tangency changes, say to \( (2, 8) \)?
3. Can you find the normal to the curve at the point \( (1, 1) \)?
4. How would you compute the tangent to an implicit curve like \( x^2 + y^2 = 1 \)?
5. What does it mean geometrically if the derivative at a point is zero?

**Tip:** The point-slope form is very useful for finding the equation of a line when you know the slope and a specific point on the line.

The tangent to y = x^3 touches the curve at (1, 1) and has a gradient of 3. The equation of the tangent is?

Learn how to find the equation of the tangent to the curve y = x^3 at the point (1, 1) with a gradient of 3. This solution involves key differentiation concepts and the point-slope formula, ideal for students in Grades 10-12.

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