2log cotx + 3log sinx - log c - 4logx + 7log y in logarithm
Answers
Answered by
3
This answer assumes that
log
(
x
)
=
log
e
(
x
)
, the natural logarithm.
First use the rule
log
(
a
b
)
=
b
log
(
a
)
:
y
=
log
(
sin
2
(
x
)
)
=
2
log
(
sin
(
x
)
)
From here, we need to know that
d
d
x
log
(
x
)
=
1
x
. Through the chain rule,
d
d
x
log
(
f
(
x
)
)
=
1
f
(
x
)
⋅
f
'
(
x
)
.
Then:
d
y
d
x
=
2
(
1
sin
(
x
)
)
(
d
d
x
sin
(
x
)
)
=
2
sin
(
x
)
(
cos
(
x
)
)
=
2
cot
(
x
)
Answered by
1
Answer:
Let y=(sinx)logx
Taking logs both sides,
logy=logxlogsinx
Differentiating both sides w.r.t. x
1ydydx=logx(1sinx)(cosx)+logsinx(1x) ⇒1ydydx=logxcotx+logsinxx
⇒dydx=y(logxcotx+logsinxx)
⇒dydx=(sinx)logx(logxcotx+logsinxx)
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