Find an equation of the tangent line to the curve at the given point.

y =(1+2x)^12, (0,1)

To find the equation of the tangent line to a curve at a given point, we need to determine the slope of the tangent line and then use the point-slope form of a line to write the equation.

Step 1: Find the derivative of the curve.
The derivative represents the slope of the tangent line at any point on the curve.

Given function: y = (1 + 2x)^12

To find the derivative, we can apply the chain rule of differentiation:

dy/dx = d/dx ((1 + 2x)^12)

Using the chain rule, we get:

dy/dx = 12(1 + 2x)^11 * d/dx (1 + 2x)

Simplifying the second term, we have:

dy/dx = 12(1 + 2x)^11 * 2

dy/dx = 24(1 + 2x)^11

Step 2: Find the slope at the given point.
To find the slope of the tangent line at the given point (0,1), substitute x = 0 into the derivative we just found:

dy/dx = 24(1 + 2(0))^11

dy/dx = 24(1)^11

dy/dx = 24

So the slope of the tangent line at (0,1) is 24.

Step 3: Use the point-slope form of a line.
The point-slope form of a line is given by y - y1 = m(x - x1), where (x1, y1) represents the given point on the line, and m is the slope of the line.

Using the equation y - y1 = m(x - x1), and substituting the values (0,1) and m = 24, we get:

y - 1 = 24(x - 0)

y - 1 = 24x

Finally, we can rewrite the equation in slope-intercept form by adding 1 to both sides:

y = 24x + 1

Therefore, the equation of the tangent line to the curve y = (1 + 2x)^12 at the point (0,1) is y = 24x + 1.

do this one just like the other one I worked out.