MATH 1500
Worksheet 5
1. If a projectile is thrown into the air, its height (in metres) is given by the formula
h(t) = −2t2 + 20t where t is in seconds.
(a) Find when the projectile returns to the ground.
Solution: The projectile hits the ground when h(t) = 0. That is,
0 = −2t2 + 20t = −2t(t − 10).
This implies that t = 0 or t = 10. But t = 0 corresponds to the time when
the projectile left the ground (not returning to the ground). Hence t = 10.
Therefore the projectile returns to the ground after 10 seconds.
(b) Find the velocity function.
Solution: The velocity function is v(t) = h′ (t) = −4t + 20.
(c) Find the velocity of the projectile after 3 seconds.
Solution: For t = 3, we have v(3) = −12 + 20 = 8. Hence the velocity of the
projectile after 3 seconds is 8 m/s.
(d) Find the velocity of the projectile when the projectile hits the ground.
Solution: The velocity of the projectile when the projectile hits the ground is
v(10) = −40 + 20.
Hence the projectile hits the ground with a velocity of −20 m/s.
(e) Find the time when the projectile stops rising.
Solution: It stops rising when the velocity is zero. That is −4t + 20 = 0. This
implies that t = 5. Hence the projectile stops rising after 5 seconds.
2. Find f ′ (x) the functions below. (Do not simplify your answers, and SHOW ALL STEPS)
√ √
2 4 x
7
f (x) = 2x − 2 x5 + 3 + 4 + 5ex + e2 .
x x
Solution: We have
5 −7
f (x) = 2x7 − 2x 2 + 2x−3 + 4x 2 + 5ex + e2 .
Hence
3 −9
f ′ (x) = 14x6 − 5x 2 − 6x−4 − 14x 2 + 5ex + 0.
3. Find the derivative of the following
(a) f (x) = (3ex )(2x5 + x2 )
Solution: By the product rule,
f ′ (x) = (3ex )′ · (2x5 + x2 ) + (3ex ) · (2x5 + x2 )′
= (3ex ) · (2x5 + x2 ) + (3ex ) · (10x4 + 2x).
(b) f (x) = (5x77 − 3xπ + 3)(ex + 1)
Solution: By the product rule,
f ′ (x) = (5x77 − 3xπ + 3)′ · (ex + 1) + (5x77 − 3xπ + 3) · (ex + 1)′
= (3855x76 − 3πxπ ) · (ex + 1) + (5x77 − 3xπ + 3) · (ex ).
√
ex + 2 x
(c) h(x) =
4x3 − ex
Solution: By the quotient rule,
√ √
′ (ex + 2 x)′ · (4x3 − ex ) − (ex + 2 x) · (4x3 − ex )′
h (x) =
(4x3 − ex )2
√
ex + √1x · (4x3 − ex ) − (ex + 2 x) · (12x2 − ex )
=
(4x3 − ex )2
(d)
tan(x) + 1
h(x) =
csc(x) − 4ex
′ ′
Take: tan(x) = sec2 (x) and csc(x) = − csc(x) cot(x)
Solution: By the quotient rule,
(tan(x) + 1)′ · (csc(x) − 4ex ) − (tan(x) + 1) · (csc(x) − 4ex )′
h′ (x) =
(csc(x) − 4ex )2
sec2 (x) · (csc(x) − 4ex ) − (tan(x) + 1) · (− csc(x) cot(x) − 4ex )
= .
(csc(x) − 4ex )2
4. Find the derivative of the following
(a) f (x) = (ex − sin(x))(2x5 + tan(x))
Solution: By the product rule,
f ′ (x) = (ex − sin(x))′ · (2x5 + tan(x)) + (ex − sin(x)) · (2x5 + tan(x))′
= (ex − cos(x)) · (2x5 + tan(x)) + (ex − sin(x)) · (10x4 + sec2 (x)).
(b) f (x) = (5x77 − 3xπ + 3)(cos(x) + 1)
Solution: By the product rule,
f ′ (x) = (5x77 − 3xπ + 3)′ · (cos(x) + 1) + (5x77 − 3xπ + 3) · (cos(x) + 1)′
= (3855x76 − 3πxπ ) · (cos(x) + 1) + (5x77 − 3xπ + 3) · (− sin(x)).
√
sec(x) + 2 x
(c) h(x) =
4x3 − cot(x)
Solution: By the quotient rule,
√ √
′ (sec(x) + 2 x)′ · (4x3 − cot(x)) − (sec(x) + 2 x) · (4x3 − cot(x))′
h (x) =
(4x3 − cot(x))2
√
sec(x) tan(x) + √1x · (4x3 − cot(x)) − (sec(x) + 2 x) · (12x2 + csc2 (x))
=
(4x3 − cot(x))2
tan(x) + 1
(d) h(x) =
csc(x) − 4ex
Solution: By the quotient rule,
(tan(x) + 1)′ · (csc(x) − 4ex ) − (tan(x) + 1) · (csc(x) − 4ex )′
h′ (x) =
(csc(x) − 4ex )2
sec2 (x) · (csc(x) − 4ex ) − (tan(x) + 1) · (− csc(x) cot(x) − 4ex )
= .
(csc(x) − 4ex )2
√
(e) f (x) = cos3 3x2 − 5
Solution: By the chain rule,
√ √ √ ′
f ′ (x) = −3 sin 3x2 − 5 · cos2 3x2 − 5 · 3x2 − 5
√ √ 1 ′
= −3 sin 3x2 − 5 · cos2 3x2 − 5 · √ · 3x2 − 5
2 3x2 − 5
√ √ 1
= −3 sin 3x2 − 5 · cos2 3x2 − 5 · √ · (6x)
2 3x2 − 5
(f) f (x) = e(2x+4) (3x2 − 2x + 1)
Solution: By the product rule and then the chain rule,
f ′ (x) = (e(2x+4) )′ · (3x2 − 2x + 1) + e(2x+4) · (3x2 − 2x + 1)′
= e(2x+4) · (2x + 4)′ · (3x2 − 2x + 1) + e(2x+4) · (6x − 2)
= 2e(2x+4) · (3x2 − 2x + 1) + e(2x+4) · (6x − 2).
2ex sin3 (4x)
(g) f (x) =
x4 + 2
Solution: By the quotient rule and then the product rule and chain rule,
(2ex sin3 (4x))′ · (x4 + 2) − (2ex sin3 (4x)) · (x4 + 2)′
f ′ (x) =
(x4 + 2)2
(2ex sin3 (4x) + 24ex cos(4x) sin2 (4x)) · (x4 + 2) − (2ex sin3 (4x)) · (4x3 )′
= .
(x4 + 2)2
