Mathematics

Mathematics

Problem

For y=(xsin⁡xcos⁡x)tan⁡xy=\bigl(x\sin x\cos x\bigr)^{\tan x}, which of the following gives dydx\dfrac{dy}{dx}? a) (xsin⁡xcos⁡x)tan⁡x[sec⁡2x ln⁡(xsin⁡xcos⁡x)−tan⁡x (1x+cot⁡x−tan⁡x)](x\sin x\cos x)^{\tan x}\Bigl[\sec^{2}x\,\ln(x\sin x\cos x)-\tan x\,\Bigl(\dfrac{1}{x}+\cot x-\tan x\Bigr)\Bigr] b) (xsin⁡xcos⁡x)tan⁡x[sec⁡2x ln⁡(xsin⁡xcos⁡x)+tan⁡x (1x+tan⁡x−cot⁡x)](x\sin x\cos x)^{\tan x}\Bigl[\sec^{2}x\,\ln(x\sin x\cos x)+\tan x\,\Bigl(\dfrac{1}{x}+\tan x-\cot x\Bigr)\Bigr] c) (xsin⁡xcos⁡x)tan⁡x[sec⁡2x ln⁡(xsin⁡xcos⁡x)+tan⁡x (1x+cot⁡x−tan⁡x)](x\sin x\cos x)^{\tan x}\Bigl[\sec^{2}x\,\ln(x\sin x\cos x)+\tan x\,\Bigl(\dfrac{1}{x}+\cot x-\tan x\Bigr)\Bigr] d) (xsin⁡xcos⁡x)tan⁡x[sec⁡2x ln⁡(xsin⁡xcos⁡x)+tan⁡x (1x+cot⁡x+tan⁡x)](x\sin x\cos x)^{\tan x}\Bigl[\sec^{2}x\,\ln(x\sin x\cos x)+\tan x\,\Bigl(\dfrac{1}{x}+\cot x+\tan x\Bigr)\Bigr]

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