1.

Find the derivative of the following function:\(f(x)=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}\)

Answer»

Let,

\(f(x)=\frac{sin\,x+cos\,x}{sin\,x-cos\,x}\)

\((sin\,x-cos\,x)\frac{d}{dx}(sin\,x+cos\,x)-(sin\,x+cos\,x)\)

\(∴f'(x)=\frac{(sin\,x-cos\,x)\frac{d}{dx}(sin\,x+cos\,x)-(sin\,x+cos\,x)​​\frac{d}{dx}(sin\,x-cos\,x)}{(sin\,x-cos\,x)^2}\)  \(=\frac{-(sin\,x-cos\,x)(cos\,x-sin\,x)-(sin\,x+cos\,x)(sin\,x+cos\,x)}{(sin\,x-cos\,x)^2}\) \(=\frac{-sin^2\,x+2sin\,x\,cos\,x\,-cos^2\,x\,-sin^2\,x-2\,sin\,x\,cos\,x-cos^2\,x}{(sin\,x\,-cos\,x)^2}\)\(=\frac{-2\,(sin^x+cos^2x)}{(sin\,x-cos^\,x)^2}\) 

\(=\frac{-2}{(sin\,x-cos^\,x)^2}\)


\(f(x)={(sinx+cosx)\over(sinx−cosx)}\)

=>\(f(x)=-{(1+tanx)\over(1-tanx)}\)

=> \(f(x)=-{tan(pi/4+x)}\)

=>\(f'(x)=-{sec^2(pi/4+x)}\)


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