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Question
Find `"dy"/"dx"` if, y = `sqrt("x" + 1/"x")`
Solution
y = `sqrt("x" + 1/"x")`
Let u = `"x" + 1/"x"`
∴ y = `sqrt"u"`
Differentiating both sides w.r.t. u, we get
`"dy"/"dx" = "d"/"du" (sqrt"u") = 1/(2sqrt"u")`
u = `"x" + 1/"x"`
Differentiating both sides w.r.t. x, we get
`"du"/"dx" = "d"/"dx"("x" + 1/"x") = 1 - 1/"x"^2`
By chain rule, we get
`"dy"/"dx" = "dy"/"du" xx "du"/"dx" = 1/(2sqrt"u") xx (1 - 1/"x"^2)`
`= 1/(2sqrt("x" + 1/"x")) (1 - 1/"x"^2)`
∴ `"dy"/"dx" = 1/2 ("x" + 1/"x")^(-1/2)(1 - 1/"x"^2)`
Alternate Method:
y = `sqrt("x" + 1/"x")`
∴ y = `("x" + 1/"x")^(1/2)`
Differentiating both sides w.r.t.x, we get
`"dy"/"dx" = "d"/"dx"[("x" + 1/"x")^(1/2)]`
`= 1/2 ("x" + 1/"x")^(-1/2) * "d"/"dx" (1 + 1/"x")`
`"dy"/"dx" = 1/2 ("x" + 1/"x")^(-1/2) (1 - 1/"x"^2)`
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