CEGEP CHAMPLAIN - ST. LAWRENCE
201-NYA-05: Differential Calculus
Patrice Camiré
Logarithmic Differentiation
1. Find the derivative of y using logarithmic differentiation. In each case, mention if logarithmic
differentiation is necessary or simply convenient.
(a) y = xx
x
2
(k) y = xe
(b) y = xsin(3x)
(c) y = [tan(2x)]x
2
(d) y = (2x + 1)3x
x+1 x
(e) y =
x−1
(f) y = xln(x)
(g) y =
sin(2x) cos(4x)
x
x
(h) y = xx
√
2x + 1(x3 + 5)2
(i) y =
e x2
(6x − 1)5 (2x + 3)4
(j) y =
(x + 1)4 (8x + 1)3
(l) y = [sin(x)]cos(x)
√
√
6x + 1 3 3x − 1
√
(m) y =
5
20x + 7
√
6
2x − 1
(n) y = √
3
2
x 3x2 + 1
(x2 + 3)8 (3x + 2)4
(x4 + 1)2 (5x − 1)3
s
ex3 (x2 + 1)9
(p) y =
(2x − 1)4 (3x − 5)3
(o) y =
(q) y = [sec(2x)]sec(2x)
(r) y = xcot(3x)
(s) y = x2 csc(x
2)
dy
2. Find
using implicit differentiation and logarithmic differentiation. Make sure to understand
dx
why logarithmic differentiation is necessary in each case.
(a) y y = xx
(e) xx = y 2 sin(x)
(b) xy = x3 y
(f) y x =
(c) xy = y x
(d) xy = x2 + x + 1
2
x
y
(g) (xy)x = x2
(h) y sin(x) = x cos(y)
Answers
1. (a)
(b)
(c)
(d)
(e)
(f)
(g)
(h)
(i)
(j)
(k)
(l)
(m)
(n)
(o)
(p)
(q)
dy
2
= x(2 ln(x) + 1)xx
dx
dy
sin(3x)
xsin(3x)
= 3 cos(3x) ln(x) +
dx
x
dy
x sec2 (2x)
2
[tan(2x)]x
= 2x ln(tan(2x)) +
dx
tan(2x)
dy
2x
(2x + 1)3x
= 3 ln(2x + 1) +
dx
2x + 1
dy
x+1
2x
x+1 x
= ln
− 2
dx
x−1
x −1
x−1
dy
= 2 ln(x)xln(x)−1
dx
dy
1 sin(2x) cos(4x)
= 2 cot(2x) − 4 tan(4x) −
dx
x
x
dy
1
x
= ln(x) + 1 +
xx ln(x)xx
dx
x ln(x)
√
dy
2x + 1(x3 + 5)2
1
6x2
=
+ 3
− 2x
dx
2x + 1 x + 5
e x2
dy
15
4
2
12
(6x − 1)5 (2x + 3)4
=2
+
−
−
dx
6x − 1 2x + 3 x + 1 8x + 1 (x + 1)4 (8x + 1)3
dy
x
= ex (x ln(x) + 1)xe −1
dx
2
dy
cos (x)
=
− sin(x) ln(sin(x)) [sin(x)]cos(x)
dx
sin(x)
√
√
dy
3
1
4
6x + 1 3 3x − 1
√
=
+
−
5
dx
6x + 1 3x − 1 20x + 7
20x + 7
√
6
dy
1
2
2x
2x − 1
√
=
− − 2
dx
3(2x − 1) x 3x + 1 x2 3 3x2 + 1
2
dy
16x
12
8x3
15
(x + 3)8 (3x + 2)4
=
+
−
−
dx
x2 + 3 3x + 2 x4 + 1 5x − 1 (x4 + 1)2 (5x − 1)3
s
1
18x
8
9
ex3 (x2 + 1)9
dy
=
−
−
3x2 + 2
dx
2
x + 1 2x − 1 3x − 5
(2x − 1)4 (3x − 5)3
dy
= 2 tan(2x)(1 + ln(sec(2x)))[sec(2x)]1+sec(2x)
dx
necessary
necessary
necessary
necessary
necessary
necessary
convenient
necessary
convenient
convenient
necessary
necessary
convenient
convenient
convenient
convenient
necessary
cot(3x)
2
− 3 csc (3x) ln(x) xcot(3x)
x
dy
1
2
2
2
(s)
= 2 csc(x )
− 2x cot(x ) ln(x) x2 csc(x )
dx
x
dy
(r)
=
dx
2. (a)
(b)
ln(x) + 1
dy
=
dx
ln(y) + 1
dy
y(3 − y)
=
dx
x(y ln(x) − 1)
dy
y(x ln(y) − y)
=
dx
x(y ln(x) − x)
1
dy
2x + 1
y
(d)
=
−
2
dx
x + x + 1 x ln(x)
(c)
(e)
necessary
necessary
dy
y
= (1 + ln(x) − cot(x))
dx
2
dy
y(1 − 2x2 ln(y))
=
dx
x(x2 + 1)
dy
y 2
(g)
=
− ln(xy) − 1
dx
x x
(f)
(h)
dy
y(1 − x cos(x) ln(y))
=
dx
x(sin(x) + y tan(y))