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Question
Find `"dy"/"dx"`, if y = `x^tanx + sqrt((x^2 + 1)/2)`
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Solution
Given that: y = `x^tanx + sqrt((x^2 + 1)/2)`
Let u = `x^tanx` and v = `sqrt((x^2 + 1)/2)`
∴ y = u + v
Differentiating both sides w.r.t. x
`"dy"/"dx" = "du"/"dx" + "dv"/"dx"` .....(i)
Now taking u = `x^tanx`
Taking log on both sides log u = `log(x^tanx)`
log u = tan x . log x
Differentiating both sides w.r.t. x
`1/"u" * "du"/"dx" = "d"/"dx"(tan x * log x)`
⇒ `1/"u" * "du"/"dx" = tan x * "d"/"dx" (log x) + log x * "d"/"dx" (tan x)`
⇒ `1/"u" * "du"/"dx" = tan x * 1/x + log x * sec^2x`
⇒ `"du"/"dx" = "u"[tanx/x + log x * sec^2x]`
∴ `"du"/"dx" = x^tanx [tanx/x + log x sec^2x]`
Taking v = `sqrt((x^2 + 1)/2)`
⇒ v = `1/sqrt(2) sqrt(x^2 + 1)`
Differentiating both sides w.r.t. x
`"dv"/"dx" = 1/sqrt(2) * 1/(2sqrt(x^2 + 1)) * 2x`
= `x/(sqrt(2)sqrt(x^2 + 1))`
Putting the values of `"du"/"dx"` and `"dv"/"dx"` in equation (i)
`"dy"/"dx" = x^tanx [log x sec^2x + tanx/x] + x/(sqrt(2)sqrt(x^2 + 1))`
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