Continuity and Differentiability — Maths STD 12 Science — Question
Gujarat 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 \quad \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 \quad \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 \quad \frac { 1 } { y } \frac { d y } { d x } = ( \cos x + \sin x )+(\cos x + \sin x) + \log(\sin x - \cos x)$
$ \Rightarrow \quad \frac { d y } { d x }= y(\cos x + \sin x)[ 1+\log(\sin x - \cos x)]$
$ \therefore \quad \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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