A quick review of some of the main points and techniques: d dx ax

A quick review of some of the main points and techniques:
d x
a = ax ln a
dx
d u
du
a = au ln a
dx
dx
loga (x) =
ln(x)
ln(a)
Problem 30: Find y 0 given
y = log5
r
(
7x ln 5
)
3x + 2
Solution:
y = log5
r
(
7x ln 5
7x ln 5
ln 5
7x
)
= log5 (
) 2 =(
) log5 (
)
3x + 2
3x + 2
2
3x + 2
7x
=(
ln 5 ln( 3x+2 )
1
7x
)
= ( ) ln(
)
2
ln(5)
2
3x + 2
y0 =
1 1
3
1
( −
)=
2 x 3x + 2
x(3x + 2)
Problem: Find y 0 given y = (sin x)cos x
Solution:
ln(y) = ln((sin x)cos x ) = cos x ln(sin x)
1
1 0
y = cos x(
) cos x + ln(sin x)(− sin x)
y
sin x
y 0 = y cos x(
1
1
)(cos x + ln(sin x)(− sin x) = (sin x)cos x (cos x(
) cos x + ln(sin x)(− sin x))
sin x
sin x
Find
Z
10
1
10
log 10(10x)
dx
x
Answer:
First note that
log10x = log10 (10) + log10 x = 1 +
Z
10
1
10
log10 (10x)
dx =
x
Z
10
(
1
10
ln x
ln(10)
1 ln x
1 ln2 x x=10
1
+
) dx = (ln(x) +
)|x= 1
10
x ln(10) x
ln(10) 2
= ln(10) − ln(10−1 ) +
= 2 ln(10)) +
ln2 (10) ln2 (10−1 )
1
(
−
)
ln(10)
2
2
1
(0) = ln(100)
ln(10)