Question

Find an equation of the tangent line to the curve at the given point: y=(1+2x)^10, (0,1)

Answer 1

Easiest way to do it. Differentiate the y equation to get y', the equation for the function's slope. The coordinates (0,1) give you values for (x1, y1), which are important for forming a point-slope equation at that coordinate.

y(x) = (2x + 1)^10
y'(x) = (10(2x + 1)^9)(2) = 20(2x + 1)^9

Now, since you already have coordinate values for x1 and y1, simply plug them into the derivative to find the slope at that coordinate.

y'(x) = 20(2x + 1)^9
y'(0) = 20(2(0) + 1)^9
y'(0) = 20(1)^9
y'(0) = 20(1)
y'(0) = 20
m = 20

Now that we have a value for the slope at x = 0 (the value defined by the given coordinate value), we can plug in all of our determined and given values into a simple point-slope form equation to finalize our slope formula for the tangent line of the given function at (0, 1).

y - y1 = m(x - x1)
m = 20
(x1, y1) = (0, 1)

y - 1 = 20(x)
y - 1 = 20x
OR
y = 20x + 1

Answer 2

Thank you! I was way over thinking this, haha.