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प्रश्न
Differentiate the following w.r.t. x :
(sin x)tanx + (cos x)cotx
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उत्तर
Let y = (sin x)tanx + (cos x)cotx
Put u = (sin x)tanx and v = (cos x)cotx
Then y = u + v
∴ `"dy"/"dx" = "du"/"dx" + "dv"/"dx"` ...(1)
Take u = (sin x)tanx
∴ log u = log(sin x)tanx = (tan x).(log sinx)
Differentiating both sides w.r.t. x, we get
`1/u."du"/"dx" = "d"/"dx"[(tan x)(log sin x)]`
= `(tan x)."d"/"dx"(log sin x) + (log sinx)."d"/"dx"(tanx)`
= `(tanx)/(sin x)."d"/"dx"(sin x) + (log sinx)(sec^2x)`
= `((sinx)/(cosx))/(sinx).cosx + (sec^2x)(log sinx)`
= 1 + (sec2x)(log sinx)
∴ `"du"/"dx" = y[1 + (sec^2x)(log sinx)]`
= (sin x)tanx[1 + (sec2x)(log sinx)] ...(2)
Also, v = (cos x)cotx
∴ log v = log(cos x)cotx = (cot x).(log cosx)
Differentiating both sides w.r.t. x, we get
`1/v."dv"/"dx" = "d"/"dx"[(cot x).(log cos x)]`
= `(cot x)."d"/"dx"(log cos x) + (log cos x)."d"/"dx"(cotx)`
= `cot x xx 1/cosx."d"/"dx"(cosx) + (log cosx).(-"cosec"^2x)`
= `cotx xx 1/cosx xx (-sin x) - ("cosec"^2x)(log cosx)`
∴ `"dv"/"dx" = v[1/tanx xx (-tanx) - ("cosec"^2x)(log cosx)]`
= –(cos x)cotx [1 + (cosec2x)(log cosx)] ...(3)
From (1), (2) and (3), we get
`"dy"/"dx" = (sin x)^(tanx)[1 + (sec^2x)(log sin x)] - (cos x)^(cotx)[1 + ("cosec"^2x)(log cosx)]`.
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