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Question
Find `"dy"/"dx"`, if x = `("u" + 1/"u")^2, "y" = (2)^(("u" + 1/"u"))`
Solution
x = `("u" + 1/"u")^2` .....(i)
Differentiating both sides w.r.t. u, we get
`"dx"/"du" = 2("u" + 1/"u") * "d"/"dx" ("u" + 1/"u")`
`= 2("u" + 1/"u") [1 + (- 1) "u"^(-2)]`
∴ `"dx"/"du" = 2("u" + 1/"u")(1 - 1/"u"^2)`
`"y" = (2)^(("u" + 1/"u"))` .....(ii)
Differentiating both sides w.r.t. u, we get
`"dy"/"du" = 2^(("u" + 1/"u")) log 2 "d"/"dx" ("u" + 1/"u")`
∴ `"dy"/"du" = log 2 * 2^(("u" + 1/"u")) (1 - 1/"u"^2)`
∴ `"dy"/"dx" = (("dy"/"du"))/(("dx"/"du")) = (2^(("u" + 1/"u")) log 2 (1 - 1/"u"^2))/(2("u" + 1/"u")(1 - 1/"u"^2))`
`= (2^(("u" + 1/"u")) log 2)/(2("u" + 1/"u"))`
∴ `"dy"/"dx" = ("y" log 2)/(2sqrt"x")` ....[From (i) and (ii)]
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