Math 1000 Tutorial
Take-Home Quiz T01 Solutions
Week 4
(1/2) 1. State the product, quotient and chain rule.
0
(a) (f (x)g(x)) =
0
Solution: (f (x)g(x)) = f 0 (x)g(x) + f (x)g 0 (x)
(b)
f (x)
g(x)
0
=
Solution:
f (x)
g(x)
0
=
f 0 (x)g(x) − f (x)g 0 (x)
[g(x)]2
0
(c) (f (g(x))) =
0
Solution: (f (g(x))) = f 0 (g(x))g 0 (x)
(1)
0
2. Prove (sec x) = sec x tan x using the rules of derivatives. [Hint: Similar to the proof for the derivative
of tan x on p. 194.]
Solution:
0
1
(sec x) =
cos x
0 · cos x − 1(− sin x)
(by quotient rule)
=
cos2 x
sin x
=
cos2 x
sin x 1
=
cos x cos x
= tan x sec x
0
(3)
3. Find the derivatives for
(a) f (x) = 14x cos x
Solution: f 0 (x) = 14 cos x + 14x(− sin x) by product rule
(b) g(x) =
x
x2 + 1
Solution: g 0 (x) =
(c) h(x) = x3 + x2 + 4
1(x2 +1)−x(2x)
(x2 +1)2
=
4. Find the derivatives for
by quotient rule
17
Solution: h0 (x) = 17 x3 + x2 + 4
(3)
−x2 +1
(x2 +1)2
16
(3x2 + 2x) by chain rule
(a) f (x) = x2 ex cos x
Solution: Let g(x) = x2 and h(x) = ex cos x, then f (x) = g(x)h(x) and thus f 0 (x) =
g 0 (x)h(x) + g(x)h0 (x) by product rule.
g 0 (x) = 2x and by product rule h0 (x) = ex cos x + ex (− sin x), therefore
f 0 (x) = 2xex cos x + x2 (ex cos x − ex sin x)
(b) g(x) =
x2
1 + x+3
ex
Solution: Let h(x) = x2 and k(x) = 1+ x+3
ex , then g(x) =
by quotient rule.
x
h(x)
k(x)
and thus g 0 (x) =
h0 (x)k(x)−h(x)k0 (x)
[k(x)]2
x
h0 (x) = 2x and by quotient rule k 0 (x) = e −(x+3)e
, therefore
e2x
x
e − (x + 3)ex
x+3
2
−x
2x 1 + x
e
e2x
g 0 (x) =
2
x+3
1+ x
e
√
(c) h(x) =
e14x+3
p
1
Solution: Let f (x) = (x) = x 2 and g(x) = e14x+3 , then h(x) = f (g(x)) and thus h0 (x) =
f 0 (g(x))g 0 (x) by chain rule.
1
f 0 (x) = 21 x− 2 and by chain rule g 0 (x) = e14x+3 (14), therefore
h0 (x) =
(1/2) 5. Find the derivative for
f (x) =
1 14x+3 − 12 14x+3
e
e
(14)
2
(x2 + 7x)esin x cos(πx − 3)
.
x4 + x + 1
Solution: Let g(x) = (x2 + 7x)esin x cos(πx − 3) and h(x) = x4 + x + 1, then f (x) =
0
f (x) =
g 0 (x)h(x)−g(x)h0 (x)
[h(x)]2
g(x)
h(x)
and thus
by quotient rule.
0
We now need to find g (x) and h0 (x). h0 (x) = 4x3 + 1 as usual. Finding the derivative for g(x) is
similar to Question 4(a):
Let m(x) = x2 + 7x and n(x) = esin x cos(πx − 3). m0 (x) = 2x + 7 and
0
0
n0 (x) = esin x cos(πx − 3) + esin x (cos(πx − 3)) by product rule
= esin x cos x cos(πx − 3) + esin x (− sin(πx − 3)π) by chain rule
Thus
g 0 (x) = (2x + 7) esin x cos(πx − 3) + (x2 + 7x) esin x cos x cos(πx − 3) + esin x (− sin(πx − 3)π)
and
0
f (x) =
(2x + 7)esin x cos(πx − 3) + (x2 + 7x) esin x cos x cos(πx − 3) − esin x sin(πx − 3)π (x4 + x + 1)
(x4 + x + 1)
−
(x2 + 7x)esin x cos(πx − 3) 4x3 + 1
You achieved a total of
(x4 + x + 1)
2
2
out of 8. Your entered grade is
out of 4 points.