Question

find the derivative of sinx+xcosx the whole divided by sinx-cosx with respect to x

Answer 1

As we know d(f/g)/dx = (g(df/dx) - f(dg/dx))/g² ,

where f and g are functions of x.Applying this we get

$$\begin{gathered} \frac{d}{dx}\left(\frac{\sin x+\cos x}{\sin x-\cos x}\right)=\frac{(\sin x-\cos x){\displaystyle \frac{d}{dx}}(\sin x+\cos x)-(\sin x+\cos x){\displaystyle \frac{d}{dx}}(\sin x-\cos x)}{(\sin x-\cos x{)}^2} \\ =\frac{(\sin x-\cos x)({\displaystyle \frac{d\left(\sin x\right)}{dx}}+{\displaystyle \frac{d\left(\cos x\right)}{dx}})-(\sin x+\cos x)({\displaystyle \frac{d\left(\sin x\right)}{dx}}-{\displaystyle \frac{d\left(\cos x\right)}{dx}})}{(\sin x-\cos x{)}^2} \\ =\frac{(\sin x-\cos x)(\cos x-\sin x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x{)}^2} \\ =\frac{-{\sin }^{2}x-{\cos }^{2}x+2\cos x\sin x-(\hspace{0.17em}{\sin }^{2}x+\hspace{0.17em}{\cos }^{2}x+2\sin x\cos x)}{(\sin x-\cos x{)}^2} \\ =\frac{-{\sin }^{2}x-{\cos }^{2}x+2\cos x\sin x-\hspace{0.17em}{\sin }^{2}x-\hspace{0.17em}{\cos }^{2}x-2\sin x\cos x)}{(\sin x-\cos x{)}^2} \\ =\frac{-2{\sin }^{2}x-2{\cos }^{2}x}{(\sin x-\cos x{)}^2} \\ =\frac{-2({\sin }^{2}x+{\cos }^{2}x)}{(\sin x-\cos x{)}^2} \end{gathered}$$