151 WebCalc Fall 2002-copyright Joe Kahlig 1
In Class Questions MATH 151-Fall 02 September 24
WebCalc gives this formula for the derivative of a function at the value x = a .
df dx x = a
= f
0
( a ) = lim h → 0 f ( a + h )
− f ( a ) h
This concept can be expressed as an alternative formula.
d dx f ( x ) x = a
= f
0
( a ) = lim a f ( x )
− f ( a ) x
− a
If you look closely at the second formula, you should notice that if you use the substitution x = a + h then the formulas are identical. Note: saying x
→ a is the same as saying h
→
0. The substitution of x = a + h should not seem that strange. This is because in both methods we are calculating the slope of the secant line and moving the second point closer to the first point.
1. Find f
0
(2) for f ( x ) = x
2 f
0
(2) = lim x → 2 f ( x ) x
−
− f
2
(2)
−
8 x + 9.
x
2
= lim x → 2
−
8 x + 9
−
(
−
3) x
−
2
= lim x → 2 x
2 −
8 x + 12 x
−
2
= lim x → 2
( x
−
2)( x
−
6) x
−
2
= lim x → 2 x
−
6 =
−
4
2. True or False: If a function is continuous at a point, then it is differentiable at that point.
This is false. The function y =
| x
| is continuous at x = 0 but is not differentiable at that point.
3. Use the definition of the derivative to compute the derivative of these functions.
(a) f f
0
( x ) =
3 x
2
+ 4 f ( x + h )
− f ( x )
( x ) = lim h → 0 h
= lim h → 0
3
( x + h ) 2 +4 h
− 3 x 2 +4
= lim h → 0 lim h → 0
3( x 2 +4)
( x 2 +2 xh + h 2 +4)( x 2 +4) h ( x
2
−
6 xh
−
3 h
2
+ 4)( x
2
+ 2 xh
− h
+
3( x 2 +2 xh + h 2 +4)
( x 2 +4)( x 2 +2 xh + h 2 +4) h
2
+ 4)
= lim h → 0 h ( x
2
= lim h → 0
+ 4)( x
2
3 x 2 +12 − 3 x 2 − 6 xh − 3 h 2 − 12
( x 2 +4)( x 2 +2 xh + h 2 +4) h
1 h (
−
6 x
−
3 h )
+ 2 xh + h
2
+ 4)
= lim h → 0 ( x
2
+ 4)(
(
−
6 x
−
3 h ) x
2
+ 2 xh + h
2
+ 4)
=
( x
2
−
6 x
+ 4)
2
(b) f ( x ) =
√
6
− x f
0 f ( x + h )
− f ( x )
( x ) = lim h → 0
√
6
− x
− h
− lim h → 0 h
√ h
6
− x
∗
= lim h → 0
√
√ p
6
−
( x + h )
−
6
− x
− h +
6
− x
− h +
√
√ h
6
− x
6
− x
=
√
6
− x lim h → 0 h (
6
−
6 x
−
− x h
−
− h
(6
+
− x )
6
− x )
= lim h → 0 h (
√
− h
6
− x
− h +
√
6
− x )
=
=
2
√
−
6
1
− x
=
4. Explain the difference between a derivative and the slope of the tangent line.
The derivative is a function. The slope of the tangent line is found by plugging in a number for x.

151 WebCalc Fall 2002-copyright Joe Kahlig 2
5. Show that the function f ( x ) =
| x
−
6
| is not differentiable at x = 6.
Remember that
| x
−
6
|
= x
−
6 for x
≥
6
−
( x
−
6) for x < 6 lim x → 6 + x
−
6
−
0 x
−
6
= lim x → 6 +
1 = 1 lim x → 6 −
−
( x
−
6)
−
0 x
−
6
= lim x → 6 −
−
1 =
−
1
Since the left hand limit and the right hand limit are not equal, the function does not have a derivative at x = 6.