Use the limit definition of the derivative to find the slope of the tangent line to the curve f(x)= 3x^2+4x+6 at x=3.

Answer 1

f '(x) = lim_{h--> 0} [f(x+h) - f(x)]/h where h is just the change in x. Basically, this is just finding the change in y over the change in x (slope) at a single point and since we're finding it at a single point there's no change so we have to take the limit as the change in the function and change in x become infinitesimally small. For these infinitesimally small changes in y and x we denote then dy and dx, so you'll see the derivative is dy/dx just the change in y over change in x but at a specific point. (Watch 3blue1brown calculus playlist especially 'the derivative paradox' for intuition).

f(x+h) = 3(x+h)² + 4(x+h) + 6 = 3x² + 6hx + 3h² + 4x + 4h + 6

f(x) = 3(x)² + 4(x) + 6

f(x+h) - f(x) = (3h² + 6hx + 4h) = h(3h + 6x + 4)

f '(x) = lim_{h--> 0} [f(x+h) - f(x)]/h = lim_{h--> 0} [h(3h + 6x + 4)]/h

h cancels out and you get 3h + 6x + 4

now we take the limit as h approaches 0, plug in h = 0 and you get 6x+4.

this is the derivative (the function that gives the slope at every point of a graph), so plug in x = 3 to get 22.