Question:
Differentiate w.r.t. x the function:\((sinx-cosx)^{(sinx-cosx)},\frac{π}{4}<x<\frac{3π}{4}\)
Differentiate w.r.t. x the function:\((sinx-cosx)^{(sinx-cosx)},\frac{π}{4}<x<\frac{3π}{4}\)
Updated On: Oct 19, 2023
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Solution and Explanation
The correct answer is \((sinx-cosx)^{(sinx-cosx)}(cosx+sinx)[1+log(sinx-cosx)]\)
Let \(y=(sinx-cosx)^{(sinx-cosx)}\)
Taking logarithm on both the sides,we obtain
\(log\,y=log[(sinx-cosx)^{(sinx-cosx)}]\)
\(⇒log\,y=(sin\,x-cos\,x).log(sin\,x-cos\,x)\)
Differentiating both sides with respect to \(x\), we obtain
\(\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[(sin\,x-cos\,x)log(sinx-cosx)]\)
\(⇒\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).\frac{d}{dx}(sinx-cosx)+(sinx-cosx).\frac{d}{dx}log(sinx-cosx)\)
\(⇒\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).(cosx+sinx)+(sinx-cosx).\frac{1}{(sinx-cosx)}.\frac{d}{dx}(sinx-cosx)\)
\(⇒\frac{1}{y}\frac{dy}{dx}=(sinx-cosx)^{(sinx-cosx)}[(cosx+sinx).log(sinx-cosx)+(cosx+sinx)]\)
\(∴\frac{dy}{dx}=(sinx-cosx)^{(sinx-cosx)}(cosx+sinx)[1+log(sinx-cosx)]\)
Let \(y=(sinx-cosx)^{(sinx-cosx)}\)
Taking logarithm on both the sides,we obtain
\(log\,y=log[(sinx-cosx)^{(sinx-cosx)}]\)
\(⇒log\,y=(sin\,x-cos\,x).log(sin\,x-cos\,x)\)
Differentiating both sides with respect to \(x\), we obtain
\(\frac{1}{y}\frac{dy}{dx}=\frac{d}{dx}[(sin\,x-cos\,x)log(sinx-cosx)]\)
\(⇒\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).\frac{d}{dx}(sinx-cosx)+(sinx-cosx).\frac{d}{dx}log(sinx-cosx)\)
\(⇒\frac{1}{y}\frac{dy}{dx}=log(sinx-cosx).(cosx+sinx)+(sinx-cosx).\frac{1}{(sinx-cosx)}.\frac{d}{dx}(sinx-cosx)\)
\(⇒\frac{1}{y}\frac{dy}{dx}=(sinx-cosx)^{(sinx-cosx)}[(cosx+sinx).log(sinx-cosx)+(cosx+sinx)]\)
\(∴\frac{dy}{dx}=(sinx-cosx)^{(sinx-cosx)}(cosx+sinx)[1+log(sinx-cosx)]\)
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