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Question
Find `dy/dx` if, y = `sqrt(x + 1/x)`
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Solution
y = `sqrt(x + 1/x)`
Let u = `x + 1/x`
∴ y = `sqrt u`
Differentiating both sides w.r.t. u, we get
`dy/(du) = 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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