Practice Problems for Differentiation Test
Math 122(N. Mackey)
Reminder: No calculators. No partial credit.
Advice: Do all the intermediate steps. Practise, practise, practise!
In each case, find an expression for the derivative.
11
1. y = √ 2
2 x −x+9
First, rewrite y in preparation for differentiation.
11
11
= (x2 − x + 9)−1/2
y= √ 2
2
2 x −x+9
Now differentiate.
−3/2
11 −1 2
dy
=
(x2 − x + 9)0
x −x+9
dx
2
2
−3/2
−11 2
=
(2x + 1)
x −x+9
4
√
2. y =
w3
w
−1
Using the quotient rule,
(w3 − 1)(w1/2 )0 − (w3 − 1)0 w1/2
(w3 − 1)2
(w3 − 1) 12 w−1/2 − 3w2 w1/2
=
(w3 − 1)2
y0 =
Multiply numerator and denominator by
y0 =
√
3. y =
w
3
w −1
√
w. Use laws of exponents. This yields
−(5w3 + 1)
√
2 w(w3 − 1)2
!7
Using logarithmic differentiation, we first find ln y
w1/2
ln y = 7 ln
w3 − 1
1
3
= 7
ln w − ln(w − 1)
2
!
(1)
Now differentiate both sides.
1 0
1
3w2
·y = 7
− 3
y
2w w − 1
#
√ !7 "
1
3w2
w
0
y = 7
−
w3 − 1
2w w3 − 1
"
#
2
4. y = ex + xe + e
2
y 0 = ex · 2x + e · xe−1 + 0
2
= 2x ex + exe−1
θ2 + 1
5. y = √
3
θ2 + 3
Taking natural log of both sides,
ln y = ln(θ2 + 1) −
1
ln(θ2 + 3)
3
Differentiating
1 0
2θ
2θ
·y = 2
−
2
y
θ + 1 3(θ + 3)
"
#
2θ(θ2 + 1)
1
1
0
y = √
−
3
θ2 + 3 θ2 + 1 3(θ2 + 3)
−1
6. y = etan
x
(x2 + 1)
Using the product rule
−1
y 0 = (etan
−1
x 0
−1
x
−1
x
) (x2 + 1) + etan
x
(x2 + 1)0
−1
· (tan−1 x)0 · (x2 + 1) + etan x (2x)
1
−1
−1
· (x2 + 1) + 2x etan x
= etan x ·
2
1+x
−1
tan−1 x
= e
+ etan x (2x)
= etan
= etan
(1 + 2x)
7. y = cos3 (5x2 )
First rewrite y in preparation for differentiation
y =
cos(5x2 )
3
2
· (cos(5x2 ))0
2
· (− sin(5x2 )) · (5x2 )0
y 0 = 3 cos(5x2 )
= 3 cos(5x2 )
= −3 cos(5x2 )
2
· sin(5x2 ) · 10x
= −30x cos2 (5x2 ) sin(5x2 )
2
8. sin(xy) = x2 − y
Use implicit differentiation.
cos(xy) · (xy)0
cos(xy) · (y + xy 0 )
cos(xy) · y + cos(xy) · (xy 0 )
y 0 [x cos(xy) + 1]
y0
2x − y 0
2x − y 0
2x − y 0
2x − y cos(xy)
2x − y cos(xy)
=
x cos(xy) + 1
=
=
=
=
9. y = e2x sec(2x) tan(2x)
Use (f gh)0 = f 0 gh + f g 0 h + f gh0 . Here,
f 0 = e2x · (2x)0 = 2e2x
g 0 = (sec(2x))0 = sec(2x) tan(2x) · (2x)0 = 2 sec(2x) tan(2x)
h0 = (tan(2x))0 = sec2 (2x) · (2x)0 = 2 sec2 (2x)
Thus,
y 0 = (e2x )0 sec(2x) tan(2x) + e2x (sec(2x))0 tan(2x) + e2x sec(2x) (tan(2x))0
= 2e2x sec(2x) tan(2x) + e2x · 2 sec(2x) tan(2x) · tan(2x) + e2x sec(2x) · 2 sec2 (2x)
h
= 2e2x sec(2x) tan(2x) + tan2 (2x) + sec2 (2x)
i
10. x2 (x − y)2 = x2 − y 2
Use implicit differentiation. The product rule applies on the LHS.
2x · (x − y)2 + x2 · 2(x − y) · (x − y)0 = 2x − 2y · y 0
2x · (x − y)2 + x2 · 2(x − y) · (1 − y 0 ) = 2x − 2y · y 0
Dividing both sides by 2,
x(x − y)2 + x2 (x − y) · (1 − y 0 ) = x − yy 0
Keep the goal in mind: we want terms involving y 0 on the LHS, all others should go the the
RHS. Use this to guide the simplification. Don’t madly multiply out everything in sight.
yy 0 − x2 (x − y)y 0 = x − x(x − y)2 − x2 (x − y)
h
i
y 0 y − x2 (x − y)
h
i
= x 1 − (x − y)2 − x(x − y)
y0 =
x [1 − (x − y)2 − x(x − y)]
y − x2 (x − y)
3
11. y =
v
u 2
u
(x
5

ln t

+ 1)13 
(2 − x3 )20
Use properties of natural log to simplify the RHS.
y =
i
1h
13 ln(x2 + 1) − 20 ln(2 − x3 )
5
Now differentiate,
1
1
1
· 2x − 20 ·
13 · 2
· (−3x2 )
5
x +1
2 − x3
26x
12x2
=
+
5(x2 + 1)
2 − x3
y0 =
12. y =
v
u
u
x(x + 2)3
3
t
(1 − x2 )7
Take natural log of both sides, so we can exploit properties of log.
ln y =
i
1h
ln x + 3 ln(x + 2) − 7 ln(1 − x2 )
3
Now differentiate both sides.
1 1
3
1
1 0
2 0
·y =
+
− 7
· (1 − x )
y
3 x
x+2
1 − x2
3
14x
1 1
+
+
=
3 x
x+2
1 − x2
Hence,
v
y0
13. y =
√
x ex
2 −1
u
3
14x
1u
x(x + 2)3 1
3
t
+
+
=
3 (1 − x2 )7 x
x+2
1 − x2
sin−1 (2x)
Use (f gh)0 = f 0 gh + f g 0 h + f gh0 . Here
1
1
f 0 = (x1/2 )0 = x−1/2 = √
2
2 x
g 0 = ex
h0
2 −1
2
· (x2 − 1)0 = 2xex −1
1
2
= q
· (2x)0 = √
1 − 4x2
1 − (2x)2
Thus
√
√
1
2
2
2
2
√ ex −1 sin−1 (2x) + x 2xex −1 sin−1 (2x) + x ex −1 √
2 x
1 − 4x2
√
2
2
ex −1
2 x ex −1
2
√ sin−1 (2x) + 2x3/2 ex −1 sin−1 (2x) + √
=
2 x
1 − 4x2
y0 =
4
t3 + 5
14. y = √
t t
A little preliminary simplification has a sweet payoff:
3
t3 + 5
t3
5
=
+ 3/2 = t3− 2 + 5t−3/2 = t3/2 + 5t−3/2
3/2
3/2
t
t
t
Differentiation is now a breeze:
√
3 −5/2
3 t
15
3 1/2
0
t
+ 5 − t
=
−
y =
2
2
2
2 t5/2
y =
15. y =
5(e2t + e−2t )
3(e2t − e−2t )
Preliminary simplification can help. Multiply numerator and denominator by e2t
5 (e2t + e−2t ) e2t
·
·
3 (e2t − e−2t ) e2t
5 (e2t e2t + e−2t e2t )
·
=
3 (e2t e2t − e−2t e2t )
5 e4t + 1
=
·
3 e4t − 1
Now differentiate using the quotient rule:
y =
(e4t − 1) · (e4t + 1)0 − (e4t − 1)0 · (e4t + 1)
(e4t − 1)2
(e4t − 1) · e4t · (4t)0 − e4t (4t)0 · (e4t + 1)
·
(e4t − 1)2
4e4t [(e4t − 1) − (e4t + 1)]
·
(e4t − 1)2
4e4t (−2)
· 4t
(e − 1)2
−40e4t
=
3(e4t − 1)2
5
3
5
=
3
5
=
3
5
=
3
y0 =
·
16. tan(x3 y 2 ) = 1
Use implicit differentiation.
sec2 (x3 y 2 ) · (x3 y 2 )0 = 0
sec2 (x3 y 2 ) · (3x2 · y 2 + x3 · 2yy 0 ) = 0
2x3 yy 0 · sec2 (x3 y 2 ) = −3x2 y 2 · sec2 (x3 y 2 )
−3y
−3x2 y 2 · sec2 (x3 y 2 )
=
y0 =
2x3 y · sec2 (x3 y 2 )
2x
Last Updated: Monday 26th October, 2015,
18:55.
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