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Differentiate w.r.t. \(x\) the function in \( (\sin x - \cos x )^{\large(\sin x - \cos x )} , \large\frac {\pi}{4}\normalsize < x < \large\frac{3\pi}{4} \)

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  • $\large\frac{d}{dx}$$(\log x)=\large\frac{1}{x}$
Step 1:
Let $y= (\sin x - \cos x )^{\large(\sin x - \cos x )}$
Taking $\log$ on both sides
$\log y=\log(\sin x - \cos x )^{\large(\sin x - \cos x )}$
$\log y=\sin x-\cos x.\log(\sin x-\cos x)$
Step 2:
$\large\frac{1}{y}\frac{dy}{dx}$$=(\cos x+\sin x)\log(\sin x-\cos x)+(\sin x-\cos x)\large\frac{d}{dx}$$\log(\sin x-\cos x)$
$\qquad=(\cos x+\sin x)\log(\sin x-\cos x)+(\sin x-\cos x)\large\frac{1}{\sin x-\cos x}\frac{d}{dx}$$(\sin x-\cos x)$
$\qquad=(\cos x+\sin x)\log(\sin x-\cos x)+(\cos x+\sin x)$
$\qquad=(\sin x+\cos x)[1+\log(\sin x-\cos x)]$
$\large\frac{dy}{dx}$$=y(\sin x+\cos x)[1+\log(\sin x-\cos x)]$
$\quad=(\sin x+\cos x)(\sin x-\cos x)^{\large\sin x-\cos x}[1+\log(\sin x-\cos x)]$
answered May 14, 2013 by sreemathi.v
 

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