find dy/dx when y=lnx/x
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Answer:
Answer:
y
'
=
ln
(
x
)
ln
(
x
)
⋅
(
1
x
⋅
ln
(
ln
(
x
)
)
+
1
x
)
Explanation:
Taking the logarithm on both sides we get
ln
(
y
)
=
ln
(
x
)
⋅
ln
(
ln
(
x
)
)
differentiating with respect to
x
:
y
'
y
=
1
x
ln
(
ln
(
x
)
)
+
ln
(
x
)
⋅
1
ln
(
x
)
⋅
1
x
so we get
y
'
=
ln
(
x
)
ln
(
x
)
⋅
(
1
x
ln
(
ln
(
x
)
)
+
1
x
)
Explanation:
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pranjali26:
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