Find the equation of the tangent line to the curve y=e^2x + 4x at the point (0,1)

dy/dx = 2 e^(2x) + 4

at the point (0,1)
dy/dx = 2e^0 + 4
= 6

y = 6x + 1

since (0,1) is the y-intercept we can just state the answer.

btw, one would expect a student to correctly spell the subject he/she is studying.

To find the equation of the tangent line to a curve at a given point, we need to determine both the slope of the tangent line and the point of tangency.

To find the slope of the tangent line at a specific point on the curve, we can take the derivative of the function representing the curve. In this case, we have the equation y = e^(2x) + 4x.

Differentiating with respect to x, we get:
dy/dx = d/dx(e^(2x)) + d/dx(4x)

The derivative of e^(2x) with respect to x can be found using the chain rule, which states that if f(x) = g(h(x)), then the derivative of f(x) with respect to x is g'(h(x)) * h'(x). In this case, g(u) = e^u and h(x) = 2x. Therefore, the derivative of e^(2x) with respect to x is e^(2x) * 2.

The derivative of 4x with respect to x is simply 4.

Therefore, dy/dx = e^(2x) * 2 + 4.

To find the slope of the tangent line at the point (0,1), we substitute x = 0 into the derivative:

dy/dx = e^(2*0) * 2 + 4 = 2 + 4 = 6.

So, at x = 0, the slope of the tangent line is 6.

Now we have the slope of the tangent line, but we still need to determine the point of tangency. For that, we use the given point (0,1).

Now that we have both the slope of the tangent line (6) and the point of tangency (0,1), we can use the point-slope form of a linear equation to find the equation of the tangent line. The point-slope form is given by the equation y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

Plugging in the values, we have:
y - 1 = 6(x - 0)

Simplifying, we get:
y - 1 = 6x

Finally, we can rewrite the equation in slope-intercept form (y = mx + b) by adding 1 to both sides:
y = 6x + 1

Therefore, the equation of the tangent line to the curve y = e^(2x) + 4x at the point (0,1) is y = 6x + 1.