Question
Differentiate the function (sin x – cos x(sin x – cos x), $\frac{\pi}{4}<x<\frac{3 \pi}{4}$ w.r.t to x.
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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Verify that (adjoint A) A = |A|I = A (adjoint A) for the above matrices.