Logarithmic Differentiation 329
Differentiating both sides with respect to x gives:
1
y
d y
dx
= (x)
1
x + 2
+ [ln(x + 2)](1),
by the product rule.
Hence
d y
dx
= y
x
x + 2
+ ln(x + 2)
= (x + 2)
x
x
x + 2
+ ln (x + 2)
When x = −1,
dy
dx
= (1)
−1
−1
1
+ ln 1
= (+1)(−1) = −1
Problem 7. Determine (a) the differential
coefficient of y =
x
√
(x − 1) and (b) evaluate
d y
dx
when x = 2.
(a) y =
x
√
(x − 1) = (x − 1)
1
x , since by the laws of
indices
n
√
a m = a
m
n
Taking Napierian logarithms of both sides gives:
ln y = ln(x − 1)
1
x =
1
x
ln(x − 1),
by law (iii) of Section 31.2.
Differentiating each side with respect to x gives:
1
y
d y
dx
=
1
x
1
x − 1
+ [ln(x − 1)]
−1
x 2
,
by the product rule.
Hence
d y
dx
= y
1
x(x − 1)
−
ln(x − 1)
x 2
i.e.
dy
dx
=
x
√
(x − 1)
1
x(x − 1)
−
ln(x − 1)
x 2
(b) When x = 2,
dy
dx
=
2
√
(1)
1
2(1)
−
ln(1)
4
= ±1
1
2
− 0
= ±
1
2
Problem 8. Differentiate x 3x+2 with respect to x.
Let y = x
3x+2
Taking Napierian logarithms of both sides gives:
ln y = ln x
3x+2
i.e. ln y = (3x + 2) ln x, by law (iii) of Section 31.2.
Differentiating each term with respect to x gives:
1
y
d y
dx
= (3x + 2)
1
x
+ (ln x)(3),
by the product rule.
Hence
d y
dx
= y
3x + 2
x
+ 3 ln x
= x 3x+2
3x + 2
x
+ 3 ln x
= x 3x+2
3 +
2
x
+ 3 ln x
Now try the following exercise
Exercise 133 Further problems on
differentiating [ f (x)]
x type functions
In Problems 1 to 4, differentiate with respect to x.
1. y = x 2x
[2x 2x (1 + ln x)]
2. y = (2x − 1) x
(2x − 1) x
2x
2x − 1
+ ln(2x − 1)
3. y =
x
√ (x + 3)
x
√
(x + 3)
1
x(x + 3)
−
ln(x + 3)
x 2
4. y = 3x 4x+1
3x 4x+1
4 +
1
x
+ 4 ln x
5. Show that when y = 2x x and x = 1,
d y
dx
= 2.
6. Evaluate
d
dx
x
√
(x − 2)
when x = 3.
1
3
7. Show that if y = θ θ and θ = 2,
d y
dθ
= 6.77,
correct to 3 significant figures.
Differentiating both sides with respect to x gives:
1
y
d y
dx
= (x)
1
x + 2
+ [ln(x + 2)](1),
by the product rule.
Hence
d y
dx
= y
x
x + 2
+ ln(x + 2)
= (x + 2)
x
x
x + 2
+ ln (x + 2)
When x = −1,
dy
dx
= (1)
−1
−1
1
+ ln 1
= (+1)(−1) = −1
Problem 7. Determine (a) the differential
coefficient of y =
x
√
(x − 1) and (b) evaluate
d y
dx
when x = 2.
(a) y =
x
√
(x − 1) = (x − 1)
1
x , since by the laws of
indices
n
√
a m = a
m
n
Taking Napierian logarithms of both sides gives:
ln y = ln(x − 1)
1
x =
1
x
ln(x − 1),
by law (iii) of Section 31.2.
Differentiating each side with respect to x gives:
1
y
d y
dx
=
1
x
1
x − 1
+ [ln(x − 1)]
−1
x 2
,
by the product rule.
Hence
d y
dx
= y
1
x(x − 1)
−
ln(x − 1)
x 2
i.e.
dy
dx
=
x
√
(x − 1)
1
x(x − 1)
−
ln(x − 1)
x 2
(b) When x = 2,
dy
dx
=
2
√
(1)
1
2(1)
−
ln(1)
4
= ±1
1
2
− 0
= ±
1
2
Problem 8. Differentiate x 3x+2 with respect to x.
Let y = x
3x+2
Taking Napierian logarithms of both sides gives:
ln y = ln x
3x+2
i.e. ln y = (3x + 2) ln x, by law (iii) of Section 31.2.
Differentiating each term with respect to x gives:
1
y
d y
dx
= (3x + 2)
1
x
+ (ln x)(3),
by the product rule.
Hence
d y
dx
= y
3x + 2
x
+ 3 ln x
= x 3x+2
3x + 2
x
+ 3 ln x
= x 3x+2
3 +
2
x
+ 3 ln x
Now try the following exercise
Exercise 133 Further problems on
differentiating [ f (x)]
x type functions
In Problems 1 to 4, differentiate with respect to x.
1. y = x 2x
[2x 2x (1 + ln x)]
2. y = (2x − 1) x
(2x − 1) x
2x
2x − 1
+ ln(2x − 1)
3. y =
x
√ (x + 3)
x
√
(x + 3)
1
x(x + 3)
−
ln(x + 3)
x 2
4. y = 3x 4x+1
3x 4x+1
4 +
1
x
+ 4 ln x
5. Show that when y = 2x x and x = 1,
d y
dx
= 2.
6. Evaluate
d
dx
x
√
(x − 2)
when x = 3.
1
3
7. Show that if y = θ θ and θ = 2,
d y
dθ
= 6.77,
correct to 3 significant figures.