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Question: If \(y = (\sin x)^{\tan x}\), then \(\frac{dy}{dx}\) is equal to...

If y=(sin⁡x)tan⁡xy = (\sin x)^{\tan x}, then dydx\frac{dy}{dx} is equal to

A

(sin⁡x)tan⁡x.(1+sec⁡2x.logsinx)(\sin x)^{\tan x}.(1 + \sec^{2}x.{logsin}x)

B

tan⁡x.(sin⁡x)tan⁡x−1.cos⁡x\tan x.(\sin x)^{\tan x - 1}.\cos x

C

(sin⁡x)tan⁡x.,sec⁡2xlogsinx(\sin x)^{\tan x}.,\sec^{2}x{logsin}x

D

tan⁡x.(sin⁡x)tan⁡x−1\tan x.(\sin x)^{\tan x - 1}

Answer

(sin⁡x)tan⁡x.(1+sec⁡2x.logsinx)(\sin x)^{\tan x}.(1 + \sec^{2}x.{logsin}x)

Explanation

Solution

Given y=(sin⁡x)tan⁡xy = (\sin x)^{\tan x}

logy=tan⁡x.logsinxy = \tan x.{logsin}x

Differentiating w.r.t. x, 1y.dydx=tan⁡x.cot⁡x+logsinx.sec⁡2x\frac{1}{y}.\frac{dy}{dx} = \tan x.\cot x + {logsin}x.\sec^{2}x

dydx=(sin⁡x)tan⁡x[1+logsinx.sec⁡2x]\frac{dy}{dx} = (\sin x)^{\tan x}\lbrack 1 + {logsin}x.\sec^{2}x\rbrack.