Continuity and Differentiability — MATHS STD 12 Science — Question
Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSContinuity and Differentiability3 Marks
Question
Differentiate the function $(\sin x – \cos x^{(\sin x – \cos x)}, \frac{\pi}{4}$
✓
Answer
We have, $y = (\sin x - \cos x)^{(\sin x - \cos x)}, \frac { \pi } { 4 } < x < \frac { 3 \pi } { 4 } $,
Therefore, on taking logarithm both sides, we get,
$\log y = \log (\sin x - \cos x)^{(\sin x - \cos x)},$
$\Rightarrow \log y= (\sin x - \cos x). \log(\sin x - \cos x)$
Therefore,on differentiating both sides $w.r.t x,$ we get,
$\frac { 1 } { y } \cdot \frac { d y } { d x } = ( \sin x - \cos x ) \times \frac { d } { d x } \log ( \sin x - \cos x ) + \log(\sin x - \cos x) \times\frac { d } { d x } ( \sin x - \cos x ) [$By using product rule of derivative$]$
$\Rightarrow \frac { 1 } { y } \frac { d y } { d x } = (\sin x - \cos x) \frac { 1 } { ( \sin x - \cos x ) } \frac { d } { d x } ( \sin x - \cos x ) + \log(\sin x - \cos x).(\cos x + \sin x)$
$\Rightarrow \frac { 1 } { y } \frac { d y } { d x } = ( \sin x - \cos x ) \frac { 1 } { ( \sin x - \cos x ) } (\cos x + \sin x) + \log(\sin x - \cos x).(\cos x + \sin x)$
$\Rightarrow \frac { 1 } { y } \frac { d y } { d x } = ( \cos x + \sin x ) +(\cos x + \sin x) + \log(\sin x - \cos x)$
$\Rightarrow \frac { d y } { d x } = y(\cos x + \sin x)[ 1+\log(\sin x - \cos x)]$
$\therefore \frac { d y } { d x } = (\sin x - \cos x)^{(\sin x - \cos x)}(\cos x + \sin x)[ 1+\log(\sin x - \cos x)]$
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