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Question
Find `"dy"/"dx"`, if x = `sqrt(1 + "u"^2), "y" = log (1 + "u"^2)`
Solution
x = `sqrt(1 + "u"^2)`
Differentiating both sides w.r.t. u, we get
`"dx"/"du" = "d"/"du"(sqrt(1 + "u"^2))`
= `1/(2sqrt(1 + "u"^2)) * "d"/"dx" (1 + "u"^2)`
= `1/(2sqrt(1 + "u"^2)) xx 2"u"`
= `"u"/sqrt(1 + "u"^2)`
y = log (1 + u2)
Differentiating both sides w.r.t. u, we get
`"dy"/"du" = "d"/"dx"[log (1 + "u"^2)]`
= `1/(1 + "u"^2) * "d"/"du" (1 + "u"^2)`
= `1/(1 + "u"^2) xx "2u"`
= `"2u"/(1 + "u"^2)`
∴ `"dy"/"dx" = (("dy"/"du"))/(("dx"/"du"))`
= `(("2u"/(1 + "u"^2)))/(("u"/sqrt(1 + "u"^2)))`
= `2/(1 + "u"^2) xx sqrt(1 + "u"^2)/"u"`
= `(2sqrt(1 + "u"^2))/(1+u^2)`
= `(2sqrt(1 + "u"^2))/(sqrt(1+u^2)xxsqrt(1+u^2))`
∴ `"dy"/"dx" = 2/sqrt(1 + "u"^2)`
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