Math 1300
Derivative Problems through 3.3
In each problem, find the derivative function.
1. f (x) = x3 · 7x
f 0 (x) = (3x2 )(7x ) + (x3 )(ln 7)(7x )
2. f (x) = 2x (4x + x2 )
f 0 (x) = (ln 2)(2x )(4x + x2 ) + (2x )(4 + 2x)
3. f (x) = ex · (2x +
√
3
x)
√
1
f 0 (x) = (ex )(2x + 3 x) + (ex )(2 + x−2/3 )
3
√
4. f (x) = e7 · x · 5x
√
1 −1/2
0
7
x
x
f (x) = e
x
(5 ) + ( x)(ln 5)(5 )
2
5. f (x) = x100 · ex
f 0 (x) = (100x99 )(ex ) + (x100 )(ex )
6. f (x) = 8 · 3x · x7
f 0 (x) = 8 (ln 3)(3x )(x7 ) + (3x )(7x6 )
Name: Solutions
Math 1300
Derivative Problems through 3.3
√
x3/2
x x
7. f (x) = x = x
2
2
x
(2 )((3/2)x1/2 ) − (x3/2 )(ln 2)(2x )
f 0 (x) =
(2x )2
x7
x2 + 1
(x2 + 1)(7x6 ) − (x7 )(2x)
f 0 (x) =
(x2 + 1)2
8. f (x) =
−3x
6x − x2
(6x − x2 )(−3) − (−3x)((ln 6)(6x ) − 2x)
f 0 (x) =
(6x − x2 )2
9. f (x) =
x+3
x−5
(x − 5)(1) − (x + 3)(1)
f 0 (x) =
(x − 5)2
10. f (x) =
x2 · 3x
3x
(3x) (2x)(3x ) + (x2 )(ln 3)(3x ) − (x2 · 3x )(3)
0
f (x) =
(3x)2
11. f (x) =
√
27x
12. f (x) =
7 + 10x
√
(7 + 10x)((2/7)x−6/7 ) − (2 7 x)(10)
f 0 (x) =
(7 + 10x)2
Name: Solutions
Math 1300
Derivative Problems through 3.3
Name: Solutions
5
5
4
4
3
3
g(x)
f (x)
13. Below are graphs of f (x) and g(x).
2
2
1
1
0
0
−1
−2
−1
0
1
−1
−2
2
−1
x
0
x
(a) Use the graphs to fill in the table with values (where they exist).
x
-1
0
1
f (x)
1
4
1
f 0 (x)
3
dne
-3
g(x)
4
1
0
g 0 (x)
-4
-2
0
(b) If
f (x)
g(x)
,
and J(x) =
,
g(x)
f (x)
fill in the following table with values (where they exist).
H(x) = f (x)g(x),
x
K(x) =
-1
0
1
H(x)
4
4
0
H 0 (x)
8
dne
0
K(x)
1/4
4
dne
K 0 (x)
1
dne
dne
J(x)
4
1/4
0
J 0 (x)
−16
dne
0
More explanations and table with computations on next page.
1
2
Math 1300
Derivative Problems through 3.3
Name: Solutions
14. The height of a rocket above the ground is given by
H = 2t t3
where H is in kilometers and t is in seconds after takeoff. Find the instantaneous velocity of
the rocket after 4 seconds.
v(t) = H 0 (t) = (ln 2)(2t )(t3 ) + (2t )(3t2 )
v(4) = (ln 2)(24 )(43 ) + (24 )(3 · 42 ) ≈ 1478 km/sec
(Okay, maybe this problem isn’t very realistic.)
15. Find the equation of the line tangent to the graph of
f (x) =
3x2
x+5
at x = 1.
3(1)2
1
f (1) =
=
1+5
2
(x
+
5)(6x)
− (3x2 )(1)
6x2 + 30x − 3x2
3x2 + 30x
f 0 (x) =
=
=
(x + 5)2
(x + 5)2
(x + 5)2
f 0 (1) =
33
11
3(1)2 + 30(1)
=
=
2
(1 + 5)
36
12
1
11
So our tangent line passes through the point 1,
and has slope
. So the equation of
2
12
the line is
11
1
y=
(x − 1) + .
12
2