http://www.ck12.org Chapter 2. Derivatives
y−y 0 =mtan(x−x 0 ).Example 1:
Find the slope of the tangent line to the curvef(x) =x^3 passing through pointP( 2 , 8 ).
Solution:
SinceP(x 0 ,y 0 ) = ( 2 , 8 ), using the slope of the tangent equation,
mtan=hlim→ 0 f(x^0 +hh)−f(x^0 )we get
mtan=limh→ 0 f(^2 +hh)−f(^2 )=limh→ 0 (h(^3) + 6 h (^2) + 12 h+ 8 )− 8
h
=limh→ 0 h
(^3) + 6 h (^2) + 12 h
h
=limh→ 0 (h^2 + 6 h+ 12 )
= 12.
Thus the slope of the tangent line is 12. Using the point-slope formula above,
y− 8 = 12 (x− 2 )
or
y= 12 x− 16
Next we are interested in finding a formula for the slope of the tangent line atany pointon the curvef. Such a
formula would be the same formula that we are using except we replace the constantx 0 by the variablex. This yields
mtan=hlim→ 0 f(x+hh)−f(x).
We denote this formula by
f′(x) =limh→ 0 f(x+hh)−f(x),
wheref(x)is read "fprime ofx." The next example illustrate its usefulness.