AP Calculus
The Derivative and the Tangent Line Problem
Bonus Problem from
Chapter 1 Test
Slope of line joining P and
O is -12/5.
Equation of line
perpendicular (tangent
line) is y +12 = 5/12 (x-5)
Let Q be a generic point
(x,y) in 4th quadrant and
calculate the slope joining
Q and P.
Solve the same problem
with a GENERIC function.
1.
Find the slope of
the secant line.
2. Then take the limit
as the secant line
approaches the
tangent line at a
defined point.
lim mx
x->5
=
If f is defined on an open interval containing c, and if:
Definition of Tangent Line
with Slope m (also called
“slope of the graph f at
exists, then the line passing through (c, f(c)) with slope m is the
x = c”
tangent line to the graph of f at the point (c, f(c)).
Example: f(x) = x2 + 1 at
the points (0,1) and (-1,2)
Vertical tangent line
Instead of doing the problem twice, once with each point, use the
definition of slope and use an arbitrary point (c, f(c)) and then
substitute the specific values in once the equation is simplified.
=
Definition of a
DERIVATIVE OF A
FUNCTION
The DERIVATIVE of f at x is given by:
f’(x) =
provided the limit exists. For all x for which this limit, exists, f’ is
a function of x.
Vocabulary and Notation
The process of finding the derivative of a function is called
differentiation.
A function is differentiable at x if its derivative exists at x and is
differentiable on an open interval (a,b), if it is differentiable at
every point in the interval.
f’(x),
, y’,
is read as “ the derivative of y with respect to x”
More examples
Find the derivative (a
function) of f(x) = x3+2x
Find the slopes of f(x) =
at (1,1), (4,2), and
(0,0)
Find the equation of the
line tangent to y = 2/t at
the point (1,2)
To find the equation of a line that is tangent to a curve, we need
the slope of that line. Therefore, we need y’, or the derivative of
the function y with respect to t, or f’(t).