1
151 WebCalc Fall 2002-copyright Joe Kahlig
In Class Questions
MATH 151-Fall 02
September 26
1. Find the derivatives of these functions.
(a) g =
3x4 + 2x2 + 6
x2
x2 (12x3 + 4x) − (3x4 + 2x2 + 6)2x
x4
g0 =
If you want to make you life a bit easier, simplify the function before taking a derivative.
g=
3x4 + 2x2 + 6
3x4 2x2
6
=
+ 2 + 2 = 3x2 + 2 + 6x−2
x2
x2
x
x
g0 = 6x − 12x−3
1
1
(b) f = 6 = x−6
2x
2
f 0 = −3x−7 or f 0 =
(c) j =
−3
x7
1
= (2x6 + 5)−1
2x6 + 5
j 0 = −1(2x6 + 5)−2 (12x5 ) =
q
(d) v =
3
−12x5
(2x6 + 5)2
1/3
(x2 + 7)(x + 5) = (x2 + 7)(x + 5)
1/3
= x3 + 5x2 + 7x + 35
−2/3 1 3
x + 5x2 + 7x + 35
3x2 + 10x + 7
3
x+7
(e) n = √
4 − x3
√
−1/2
4 − x3 (1) − (x + 7) ∗ 12 4 − x3
∗ (−3x2 )
0
√
n =
( 4 − x3 )2
Now if this was a multiple choice question, then you would have to simplify. There are
two main methods to simplify. I’ll let you choose which one you like the best.
v0 =
Method 1:
√
n0 =
√
=
4 − x3 − (x + 7) ∗ 12 4 − x3
√
( 4 − x3 )2
4 − x3 ∗
√
3
2
2∗√4−x3
√+21x
+ 3x
2∗ 4−x3
2 4−x3
(4 − x3 )
−1/2
∗ (−3x2 )
3
=
8−2x
√
2∗ 4−x3
+
(4 −
√
=
2)
(x+7)∗(−3x
√
2 4−x3
− x3 )
4 − x3 −
3 +21x2
3x√
2 4−x3
x3 )
(4
=
3 +21x2
8−2x3 +3x
√
2∗ 4−x3
4−x3
1
=
=
Method 2:
√
0
n =
=
4 − x3 − (x + 7) ∗ 12 4 − x3
√
( 4 − x3 )2
−1/2
∗ (−3x2 )
√
2 4 − x3
∗ √
=
2 4 − x3
2(4 − x3 ) − (x + 7) ∗ (−3x2 )
8 − 2x3 + 3x3 + 21x2
8 + x3 + 21x2
√
√
√
=
=
2(4 − x3 ) 4 − x3
2(4 − x3 ) 4 − x3
2(4 − x3 ) 4 − x3
8 + x3 + 21x2
√
2(4 − x3 ) 4 − x3
2
151 WebCalc Fall 2002-copyright Joe Kahlig
2. Find the equation of the tangent line at x = 3 for f (x) = (x3 − 6)5 + 28
f 0 (x) = 5(x3 − 6)4 ∗ 3x2 . To find the equation of the tangent line we need a point and a slope.
f (3) = 4084357 and f 0 (3) = 26254935
Answer: y − 4084357 = 26254935(x − 3)
3. Use the table of information to compute the following. If it is not possible explain why.
x
f (x)
f 0 (x)
2
5
6
3
4
7
4
3
8
5
2
9
x
g(x)
g0 (x)
2
1
−5
3
2
−6
4
5
−7
5
8
−9
J(x) = (2f + g)(x) = 2f (x) + g(x)
H(x) =
f
f (x)
(x) =
f −g
f (x) − g(x)
K(x) = (x2 + 7)f (x)
(a) J 0 (x) = 2f 0 (x) + g0 (x) so J 0 (4) = 2f 0 (4) + g0 (4) = 2 ∗ 8 + −7 = 16 − 7 = 9
(b) H 0 (x) =
H 0 (5) =
(f (x) − g(x)) ∗ f 0 (x) − f (x) ∗ (f 0 (x) − g0 (x))
(g(x))2
(f (5) − g(5)) ∗ f 0 (5) − f (5) ∗ (f 0 (5) − g0 (5))
(2 − 8) ∗ 9 − 2 ∗ (9 − −9)
−90
=
=
2
2
(g(5))
(8)
64
(c) K 0 (3)
df
4. Find
if the function is defined as f =
da
x9 − 6x3 + 7
√
(x3 − 6)10 x8 − 10
!10
df
The derivative of this function is zero.
tells us to take the derivative with respect to the
da
variable a.