School and Work: tangent

Open Question: Calculus: Find an equation of the tangent line to the curve at the given point?

f(x) = (1 + 2x)¹° so f '(x) = 10(1 + 2x)? •2 = 20(1 + 2x)?

Since the point is (0,1) the slope at that point is f '(0) = 20(1)? = 20

So using the point-slope form of the line

y – 1 = 20(x – 0) or y = 20x + 1

Sorry, I could not read the content fromt this page.

The slope of the tangent is where the derivative is 0. So you have to use product rule and implicit differentiation to get:

3x^2 + 3y^2 * y' = 3xy' + 3y

Solving for y':
3y^2 y' - 3xy' = 3y - 3x^2
y' [3y^2 - 3x] = 3y - 3x^2
y' = (3y - 3x^2) / (3y^2 - 3x)
y' = (y - x^2) / (y^2 - x)

When y' = 0 gives 0 = y - x^2 => At (x, y) the gradient is m = x^2/y