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Question #226792
(Cos(2x))^1/(x^2) please help with the derivative. Thanks
Expert's answer
y
=
(
cos
(
2
x
)
)
1
/
x
2
y=(\cos(2x))^{1/x^2}
y
=
(
cos
(
2
x
)
)
1/
x
2
ln
y
=
ln
(
cos
(
2
x
)
)
x
2
\ln y=\dfrac{\ln(\cos(2x))}{x^2}
ln
y
=
x
2
ln
(
cos
(
2
x
))
Differentiate both sides with respect to
x
x
x
y
′
y
=
−
2
x
2
tan
(
2
x
)
−
2
x
ln
(
cos
(
2
x
)
)
x
4
\dfrac{y'}{y}=\dfrac{-2x^2\tan(2x)-2x\ln(\cos(2x))}{x^4}
y
y
′
=
x
4
−
2
x
2
tan
(
2
x
)
−
2
x
ln
(
cos
(
2
x
))
y
′
=
−
2
x
3
(
x
tan
(
2
x
)
+
ln
(
cos
(
2
x
)
)
)
(
cos
(
2
x
)
)
1
/
x
2
y'=-\dfrac{2}{x^3}\big(x\tan(2x)+\ln(\cos(2x))\big)\big(\cos(2x)\big)^{1/x^2}
y
′
=
−
x
3
2
(
x
tan
(
2
x
)
+
ln
(
cos
(
2
x
))
)
(
cos
(
2
x
)
)
1/
x
2
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#340153
on Dec 2023
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