MATH1131Q – Extra Practice with Derivatives – Solutions
1
David Nichols
Taking Derivatives
(a)
√
f (x) = 2( x(x2 + sin x))
0
f (x) = 2
√
1
2
x(2x + cos x) + √ (x + sin x)
2 x
(b)
f (t) =
csc t
t2
f 0 (t) =
(− csc t cot t)t2 − (csc t)2t
t4
(c)
g(x) =
x + ln x
sin x cos x
g 0 (x) =
(1 + x1 )sin x cos x − (x + ln x)(cos2 x − sin2 x)
(sin x cos x)2
(d)
y=
ax cos x
x2 + 1
dy
(1 · cos x + x(− sin x))(x2 + 1) − (x cos x)2x
=a
dx
x2 + 1
(e)
We can use the chain rule, but logarithmic differentiation is (arguably) easier.
p
h(x) = 3 (x + 1)(x − 2)(x + 7)2
hp
i
ln [h(x)] = ln 3 (x + 1)(x − 2)(x + 7)2
1 ln [h(x)] = ln (x + 1)(x − 2)(x + 7)2
3
1
ln [h(x)] = [ln(x + 1) + ln(x − 2) + 2 ln(x + 7)]
3
1 0
1
1
1
1
h (x) =
+
+2
h(x)
3 x+1 x−2
x+7
1
1
2
1
0
+
+
h (x) = h(x)
3
x+1 x−2 x+7
1p
1
1
2
3
0
h (x) =
(x + 1)(x − 2)(x + 7)2
+
+
3
x+1 x−2 x+7
(f)
f (x) = tan10 (3x)
10
f (x) = tan(3x)
9
f 0 (x) = 10 tan(3x) · sec2 (3x) · 3
(g)
p
5
sec(tan x)
1/5
y = sec(tan x)
y=
−4/5
dy
1
= sec(tan x)
· sec(tan x)tan(tan x) · sec2 x
dx
5
(h)
g(t) =
g 0 (t) =
(t4
1
+ 1)6 + 1
[(t4 + 1)6 + 1] · 0 − 1 · [6(4t + 0)5 + 0]
[(t4 + 1)6 + 1]2
(i)
h(t) = sin(sin(sin t))
h0 (t) = cos(sin(sin t)) · cos(sin t) · cos t
(j)
Because the power rule only words for xconstant , we have to use logarithmic
differentiation.
h(x) = xx
2
2
ln[h(x)] = ln[xx ]
ln[h(x)] = x2 ln x
1 0
1
h (x) = 2x ln x + x2
h(x)
x
0
h (x) = h(x)(2x ln x + x)
2
h0 (x) = xx (2x ln x + x)
(k)
Whenever we have x in both the base and the exponent, we use logarithmic
differentiation to take the derivative.
f (x) = (tan(2x))x
ln[f (x)] = ln[(tan(2x))x ]
ln[f (x)] = x ln(tan(2x))
1
1 0
f (x) = 1 · ln(tan(2x)) + x ·
· sec2 (2x) · 2
f (x)
tan(2x)
2x
0
f (x) = f (x) ln(tan(2x)) +
sin(2x) cos(2x)
2x
0
x
f (x) = (tan(2x)) ln(tan(2x)) +
sin(2x) cos(2x)
(l)
√
y = cos(x 7 − x2 )
√
√
dy
1
2 −1/2
2
2
= − sin(x 7 − x ) · 1 · 7 − x + x (7 − x )
· (−2x)
dx
2
(m)
f (t) = k(t3 − tan t)3
f 0 (t) = k 3(t3 − tan t)2 · (3t2 − sec2 t)
(n)
g(x) = ex ln x
g 0 (x) = ex ln x + ex
1
x
2
Implicit Derivatives
(a)
x2 + y 2 = r 2
d 2
d 2
(x + y 2 ) =
(r )
dx
dx
dy
2x + 2y
=0
dx
dy
x
=−
dx
y
(b)
√
1
x + √ = 4x
y
d √
1
d
(4x)
x+ √
=
dx
y
dx
1 dy
1
√ − p
=4
2 x 2 y 3 dx
r
dy
y3
=4
dx
x
(c)
√
xy + x + y = 5
√
d √
d
xy + x + y =
(5)
dx
dx
dy
1
dy
1
y+x
+ √
1+
=0
√
2 xy
dx
2 x+y
dx
r
r
1
y
x dy
1
1
dy
+
+√
+√
=0
2
x
y dx
x+y
x + y dx
√
dy
=−
dx
r
x2
y
+ xy
(d)
x y
+ = xy
y x
d x y
d
+
=
(xy)
dx y x
dx
dy
dy
y − x dx
−y
x dx
dy
+
=x +y
2
2
y
x
dx
x2 y 3 − x2 y + y 3
dy
=− 3 2
dx
x y − xy 2 + x3
(e)
sin(3x) = cos(2y)
d
d
sin(3x) =
cos(2y)
dx
dx
3 cos(3x) = −2 sin(2y)
−
dy
dx
3 cos(3x)
dy
=
2 sin(2y)
dx
(f)
2x
(x2 + y 2 )1/2 = 11 − 2
(x + y 2 )1/2
d
d 2
2x
2 1/2
(x + y ) =
11 − 2
dx
dx
(x + y 2 )1/2
dy
1
2
2 1/2
2(x + y ) − 2x 2(x2 +y2 )1/2 (2x + 2y dx )
1
dy
(2x
+
2y
)
=
−
2(x2 + y 2 )1/2
dx
x2 + y 2
dy
x3 + (2y + 2)x2 + xy 2 + 2y 3
=−
dx
x2 y + 2xy + y 3
(g)
cot y − x sin x = 9
d
d
(cot y − x sin x) =
(9)
dx
dx
dy
(− csc2 y) − sin x − x cos x = 0
dx
dy
= − sin2 y(sin x + x cos x)
dx
(h)
√
x
d
d √
((2x − y)3 ) =
( x)
dx dx
dy
1
3(2x − y)2 2 −
= √
dx
2 x
(2x − y)3 =
1
dy
=2− √
dx
6 x(2x − y)2
(i)
sin(x − y) = c − y
d
d
(sin(x − y)) =
(c − y)
dx dx
dy
dy
cos(x − y) 1 −
=−
dx
dx
dy
cos(x − y)
=
dx
cos(x − y) − 1
(j)
x9 y 8 = ln x
d 9 8
d
(x y ) =
(ln x)
dx
dx
dy
1
9x8 y 8 + 8x9 y 7
=
dx
x
dy
1 − 9x9 y 8
=
dx
8x10 y 7
(k)
x3 + (x2 − 1)y 2 + y 3 + 2x = 0
d
d 3
(x + (x2 − 1)y 2 + y 3 + 2x) =
(0)
dx
dx
dy
dy
+ 3y 2
+2=0
3x2 + 2xy 2 + 2(x2 − 1)y
dx
dx
3x2 + 2xy 2 + 2
dy
=−
dx
2(x2 − 1)y + 3y 2
(l)
x tan y = y tan x
d
d
(x tan y) =
(y tan x)
dx
dx
dy
dy
tan y + x(sec2 y)
=
tan x + y sec2 x
dx
dx
dy
y sec2 x − tan y
=
dx
x sec2 y − tan x
(m)
√
1
4 √
d
d 1
( x2 − 4 + ln y) =
dx
dx 4
x
1 dy
√
+
=0
x2 − 4 y dx
x2 − 4 + ln y =
xy
dy
= −√
dx
x2 − 4
(n)
7ey + 7ex = x7
d
d 7
(7ey + 7ex ) =
(x )
dx
dx
dy
7ey
+ 7ex = 7x6
dx
dy x6 − ex
=
dx
ey
3
Higher Derivatives
(a)
f (x) = − ln(cos x)
−1
· − sin x = tan x
f 0 (x) =
cos x
f 00 (x) = sec2 x
(b)
2
f (x) = 1 + x3/2
3
2
3
f 0 (x) = 0 + · · x1/2 = x1/2
3 2
1
f 00 (x) = x−1/2
2
(c)
f (x) = x ln x − x
f 0 (x) = 1 · ln x + x ·
f 00 (x) =
1
− 1 = ln x
x
1
x
(d)
f (x) = x sin x
f 0 (x) = 1 · sin x + x · cos x = sin x + x cos x = sin x + x cos x
f 00 (x) = cos x + 1 · cos x + x · (− sin x) = 2 cos x − x sin x
(e)
√
x2 + 25
1
x
x
f 0 (x) = √
· 2x = √
=√
2 x2 + 25
x2 + 25
x2 + 25
f (x) =
00
f (x) =
1·
√
x2 + 25 − x ·
x2
+ 25
√ x
x2 +25
=
(x2
25
+ 25)3/2
(f)
ex + e−x
1
= (ex + e−x )
2
2
1
1
f 0 (x) = (ex + e−x (−1)) = (ex − e−x )
2
2
1
1
f 00 (x) = (ex − e−x (−1)) = (ex + e−x )
2
2
f (x) =
(g)
f (x) = e3x
2
2
2
f 0 (x) = e3x · 6x = 6xe3x = 6x e3x
2
2
2
f 00 (x) = 6e3x + 6x 6xe3x = (6 + 36x2 )e3x
2
(h)
f (x) = sin(ln x)
f 0 (x) = cos(ln x) ·
00
f (x) =
(i)
x)
x − sin(ln
x
1
cos(ln x)
cos(ln x)
=
=
x
x
x
− 1 · cos(ln x)
sin(ln x) + cos(ln x)
=−
2
x
x2
Refer back to (d).
f (x) = cos x + x sin x
f 0 (x) = − sin x + sin x + x cos x = x cos x
f 00 (x) = cos x − x sin x
(j)
Refer back to (d).
f (x) = 2x sin x + (2 − x2 ) cos x
f 0 (x) = 2 sin x + 2x cos x + (x2 − 2) sin x − 2x cos x = x2 sin x
f 00 (x) = 2x sin x + x2 cos x
(k)
1
f (x) = ex (sin x − cos x)
2
1
f 0 (x) = (ex (sin x − cos x) + ex (cos x + sin x)) = ex sin x = ex sin x
2
00
f (x) = ex sin x + ex cos x = ex (sin x + cos x)
(l)
f (x) = sec x
f 0 (x) = sec x tan x = sec x tan x
f 00 (x) = sec x tan x tan x + sec x sec2 x = sec x tan2 x + sec3 x
(m)
f (x) = x10 + 2x6 − 3x5 + x2
f 0 (x) = 10x9 + 12x5 − 15x4 + 2x
f 00 (x) = 90x8 + 60x4 − 60x3 + 2
(n)
x2
1 − x2
√
−x
2x 1 − x2 − x2 √1−x
2
f (x) = √
x3 − 2x
x3 − 2x
=
−
1 − x2
(1 − x2 )3/2
(1 − x2 )3/2
3x2 − 2(1 − x2 )3/2 − (x3 − 2x) 32 (1 − x2 )1/2
x2 + 2
00
f (x) = −
=
(1 − x2 )3
(1 − x2 )5/2
f 0 (x) =
=−