The equation of the tangent to the curve `y = e^(2x) ` at (0,1) is

Options

A. A) y + 1 = 2 x
B. B) 1 - y = 2 x
C. C) y - 1 = 2 x
D. D) none of these

Answer

To find the equation of the tangent to the curve \( y = e^{2x} \) at the point \( (0, 1) \), we can follow these steps:

### Step 1: Find the derivative of the function
The first step is to differentiate the function \( y = e^{2x} \) with respect to \( x \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = \frac{d}{dx}(e^{2x}) = 2e^{2x} \]

### Step 2: Evaluate the derivative at the point of tangency
Next, we need to evaluate the derivative at the point \( (0, 1) \) to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=0} = 2e^{2 \cdot 0} = 2e^{0} = 2 \cdot 1 = 2 \]

So, the slope \( m \) of the tangent line at the point \( (0, 1) \) is \( 2 \).

### Step 3: Use the point-slope form of the equation of a line
Now that we have the slope and the point, we can use the point-slope form of the equation of a line, which is given by:

\[ y - y_1 = m(x - x_1) \]

Substituting \( m = 2 \), \( x_1 = 0 \), and \( y_1 = 1 \):

\[ y - 1 = 2(x - 0) \]

### Step 4: Simplify the equation
Now, we simplify the equation:

\[ y - 1 = 2x \]

Adding \( 1 \) to both sides gives:

\[ y = 2x + 1 \]

### Conclusion
Thus, the equation of the tangent to the curve \( y = e^{2x} \) at the point \( (0, 1) \) is:

\[ y - 1 = 2x \]

### Final Answer
The correct option is \( y - 1 = 2x \).

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