d/{dx}[x^{sinx} + (sinx)^x ] =

Correct Answer is : xsinx [sinxx+cosxlogx] + (sinx)x [x tan x + log (sin x)]

x^{sinx} [{sinx}/x + cosxlogx] + (sinx)^x [x tan x + log (sin x)]

x^{sinx} [x tan x + log (sin x)] + (sinx)^x x^{sinx} [{sinx}/x + \log(sinx)]

x^{sinx} [x/{sinx} + cosxlogx] + (sinx)^x [x cot x + log (sin x)]

x^{sinx} [\frac{sinx}{x} + sinxlogx] + (sinx)^x [x cot x + log (cos x)]

AP EAMCET Engineering 2018 Apr 24 Shift 1 Paper

Section: Mathematics

Question : 66 of 160

Marks: +1, -0

[xsinx+(sinx)x] =

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