Continuity and Differentiability — Class 12 Maths Solution

ncert misc SA NCERT Misc. ,Q.9,Page 191
Question

${(\sin x - \cos x)^{(\sin x - \cos x)}},\cfrac{\pi }{4} < x < \cfrac{{3\pi }}{4}.$

Step-by-step Solution

Let $y = {(\sin x - \cos x)^{(\sin x - \cos x)}}$
By taking log on both sides , we get
log y $=$ (sin x $-$ cos x) log (sin x $-$ cos x) ...(i)

Differentiating (i) on both sides w.r.t. x, we get

$\Rightarrow$ $\cfrac{1}{y}\cfrac{{dy}}{{dx}} = (\sin x - \cos x)\cfrac{{(\cos x + \sin x)}}{{(\sin x - \cos x)}} + \log (\sin x - \cos x) \cdot (\cos x + \sin x)$

$\Rightarrow$ $\cfrac{1}{y}\cfrac{{dy}}{{dx}} = (\cos x + \sin x) + \log (\sin x - \cos x) \cdot (\cos x + \sin x)$

$\Rightarrow$ $\cfrac{{dy}}{{dx}} = y(\cos x + \sin x)[1 + \log (\sin x - \cos x)]$

therefore, $\cfrac{{dy}}{{dx}} = {(\sin x - \cos x)^{(\sin x - \cos x)}}$
$[(\cos x + \sin x)(1 + \log (\sin x - \cos x))],\sin x > \cos x$

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