differentiation

328 DIFFERENTIAL CALCULUS
i.e. dy
{
dx = x√ 1
(x − 1)
x(x − 1)
(b) When x = 2,
Problem 8.
to x.
}
ln(x − 1)
−
x 2
{ }
1 ln (1)
−
2(1) 4
{ } 1
=±1
2 − 0 = ± 1 2
dy
dx = 2√ (1)
Differentiate x 3x+2 with respect
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 dy
1
y dx = (3x + 2) + (ln x)(3),
x
Hence
{
dy 3x + 2
dx = y + 3lnx
x
= x 3x+2 { 3x + 2
x
by the product rule.
}
}
+ 3lnx
= x 3x+2 {
3 + 2 x + 3lnx }
Now try the following exercise.
Exercise 134 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√ 1
(x + 3)
x(x + 3)
−
ln(x + 3)
x 2 }]
4. y = 3x 4x+1 [
3x 4x+1 {
4 + 1 x + 4lnx }]
5. Show that when y = 2x x and x = 1, dy
dx = 2.
6. Evaluate d { √ x
}
(x − 2) when x = 3.
dx
[ ] 1
3
7. Show that if y = θ θ and θ = 2, dy
dθ =6.77,
correct to 3 significant figures.