Differentiate given problems w.r.t.x.
(sin x−cos x)(sin x−cos x)
Let y=(sin x−cos x)(sin x−cos x)
Taking log on both sides w.r.t.x, we get
log y = log(sin x−cos x)(sin x−cos x)
or logy=(sinx−cosx)log(sinx−cosx)(∴logmn=n log.m)
Differentiating both sides w.r.t.x, we obtain
1ydydx=(sinx−cosx)cos x+sin xsin x−cos x+[log(sinx−cosx)](cosx+sinx)
⇒dydx=y(cosx+sinx)[1+log(sin x−cos x)]
=(cosx+sinx)(sinx−cosx)(sinx−cosx)1+log(sinx−cosx)