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рдкреНрд░рд╢реНрди
Find `(dy)/(dx), if y = x^tanx + sqrt(x^2 + 1)/2`.
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`y = x^tanx + sqrt(x^2 + 1)/2 ...(i)`
Let u = `x^tanx, v = sqrt(x^2 + 1)/2`
Equation (i) becomes
y = u + v
Differentiate w.r.t. x,
`(dy)/(dx) = (du)/(dx) + (dv)/(dx)`
Now, u = xtan x
taking log on both sides
log u = log (xtan x)
log u = tan x log x
Differentiate w.r.t. x,
`1/u (du)/(dx) = (tan x)/x + log x sec2^2 x`
⇒ `(du)/(dx) = u[(tan x)/x + log x sec^2 x]`
⇒ `(du)/(dx) = x^tan x[(tan x)/x + log x sec^2 x]`
Again take,
v = `sqrt(x^2 - 1)/2`
`(dv)/(dx) = 1/2 xx 1/(2sqrt(x^2 + 1)) xx (2x)`
= `x/(2sqrt(x^2 + 1)`
Put the value `(du)/(dx) and (dv)/(dx) "in equation" (ii),`
Now, `(dy)/(dx) = x^tanx[tan x/x + log x sec^2 x] + x/(2sqrt(x^2 + 1)`
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