To find the equation of the tangent line to [tex]\( y = x^2 \)[/tex] at the point [tex]\((1, 1)\)[/tex], we need to follow these steps:
### Step 1: Differentiate the function
The given function is [tex]\( y = x^2 \)[/tex]. We need the derivative of [tex]\( y \)[/tex] with respect to [tex]\( x \)[/tex] to find the slope of the tangent line.
[tex]\[ \frac{dy}{dx} = \frac{d}{dx}(x^2) = 2x \][/tex]
### Step 2: Evaluate the slope at the given point
Now, we evaluate the derivative at the point [tex]\( x = 1 \)[/tex].
[tex]\[ \frac{dy}{dx} \Bigg|_{x=1} = 2 \cdot 1 = 2 \][/tex]
So, the slope of the tangent line at [tex]\( x = 1 \)[/tex] is [tex]\( 2 \)[/tex].
### Step 3: Use the point-slope form of the tangent line
The point-slope form of a line is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\( (x_1, y_1) \)[/tex] is the point of tangency and [tex]\( m \)[/tex] is the slope. Here, [tex]\( (x_1, y_1) = (1, 1) \)[/tex] and [tex]\( m = 2 \)[/tex].
[tex]\[ y - 1 = 2(x - 1) \][/tex]
### Step 4: Simplify the equation
Solve for [tex]\( y \)[/tex] to put the equation in slope-intercept form [tex]\( y = mx + b \)[/tex].
[tex]\[ y - 1 = 2x - 2 \][/tex]
[tex]\[ y = 2x - 2 + 1 \][/tex]
[tex]\[ y = 2x - 1 \][/tex]
So, the equation of the tangent line is [tex]\( y = 2x - 1 \)[/tex].
### Step 5: Verify the options
Let's verify which option matches our derived equation:
- [tex]\( y = 2x - 2 \)[/tex]
- [tex]\( y = 2x - 1 \)[/tex] ✔️
- [tex]\( y = 3x + 1 \)[/tex]
- [tex]\( y = 2x + 2 \)[/tex]
### Conclusion
The correct equation of the tangent line to [tex]\( y = x^2 \)[/tex] at [tex]\((1, 1)\)[/tex] is [tex]\( y = 2x - 1 \)[/tex]. Therefore, the correct answer is:
[tex]\[ \boxed{y = 2x - 1} \][/tex]
