if y=1/√x then dy/dx is equal to
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We have, y=x
x
Taking log on both the sides, we get
logy=xlogx
On differentiating w.r.t. x, we get
y
1
dx
dy
=
x
x
+logx
⇒
dx
dy
=y+ylogx
⇒
dx
dy
=x
x
(1+logx) ....(∵y=x
x
)
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