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Question #257106
use chain rule to find dy/dx for given value of x
y=(u-1/u+1)^1/2, u=√x-1, for x=34/9
Expert's answer
y
=
(
u
−
1
u
+
1
)
1
/
2
,
u
=
x
−
1
y=(\dfrac{u-1}{u+1})^{1/2}, u=\sqrt{x-1}
y
=
(
u
+
1
u
−
1
)
1/2
,
u
=
x
−
1
y
x
′
=
1
2
(
u
−
1
u
+
1
)
−
1
/
2
(
u
+
1
−
(
u
−
1
)
(
u
+
1
)
2
)
u
x
′
y'_x=\dfrac{1}{2}(\dfrac{u-1}{u+1})^{-1/2}(\dfrac{u+1-(u-1)}{(u+1)^2})u'_x
y
x
′
=
2
1
(
u
+
1
u
−
1
)
−
1/2
(
(
u
+
1
)
2
u
+
1
−
(
u
−
1
)
)
u
x
′
=
(
u
+
1
u
−
1
)
1
/
2
(
1
(
u
+
1
)
2
)
(
1
2
x
−
1
)
=(\dfrac{u+1}{u-1})^{1/2}(\dfrac{1}{(u+1)^2})(\dfrac{1}{2\sqrt{x-1}})
=
(
u
−
1
u
+
1
)
1/2
(
(
u
+
1
)
2
1
)
(
2
x
−
1
1
)
=
(
x
−
1
+
1
x
−
1
−
1
)
1
/
2
(
1
(
x
−
1
+
1
)
2
)
(
1
2
x
−
1
)
=(\dfrac{\sqrt{x-1}+1}{\sqrt{x-1}-1})^{1/2}(\dfrac{1}{(\sqrt{x-1}+1)^2})(\dfrac{1}{2\sqrt{x-1}})
=
(
x
−
1
−
1
x
−
1
+
1
)
1/2
(
(
x
−
1
+
1
)
2
1
)
(
2
x
−
1
1
)
x
=
34
9
:
u
=
34
9
−
1
=
5
3
x=\dfrac{34}{9}:u=\sqrt{\dfrac{34}{9}-1}=\dfrac{5}{3}
x
=
9
34
:
u
=
9
34
−
1
=
3
5
y
′
(
34
9
)
=
(
5
/
3
+
1
5
/
3
−
1
)
1
/
2
(
1
(
5
/
3
+
1
)
2
)
(
1
2
(
5
/
3
)
y'(\dfrac{34}{9})=\bigg(\dfrac{5/3+1}{5/3-1}\bigg)^{1/2}\big(\dfrac{1}{(5/3+1)^2}\big)(\dfrac{1}{2(5/3})
y
′
(
9
34
)
=
(
5/3
−
1
5/3
+
1
)
1/2
(
(
5/3
+
1
)
2
1
)
(
2
(
5/3
1
)
=
27
320
=\dfrac{27}{320}
=
320
27
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#340153
on Dec 2023
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