Question

derivative of 1-tanx/1+tanx

Answer 1

$$\begin{gathered} \mathrm{Let}\mathrm{y}=\frac{1-\tan \mathrm{x}}{1+\tan \mathrm{x}} \\ \Rightarrow \mathrm{y}=\frac{1-{\displaystyle \frac{\sin \mathrm{x}}{\cos \mathrm{x}}}}{1+{\displaystyle \frac{\sin \mathrm{x}}{\cos \mathrm{x}}}} \\ \Rightarrow \mathrm{y}=\frac{\cos \mathrm{x}-\sin \mathrm{x}}{\cos \mathrm{x}+\sin \mathrm{x}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)\times {\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\cos \mathrm{x}-\sin \mathrm{x}\right)-\left(\cos \mathrm{x}-\sin \mathrm{x}\right)\times {\displaystyle \frac{\mathrm{d}}{\mathrm{dx}}}\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}{{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)\left(-\sin \mathrm{x}-\cos \mathrm{x}\right)-\left(\cos \mathrm{x}-\sin \mathrm{x}\right)\left(-\sin \mathrm{x}+\cos \mathrm{x}\right)}{{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}^2-{\left(\cos \mathrm{x}-\sin \mathrm{x}\right)}^2}{{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\left({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}+2\sin \mathrm{x}.\cos \mathrm{x}\right)-\left({\cos }^2\mathrm{x}+{\sin }^2\mathrm{x}-2\sin \mathrm{x}.\cos \mathrm{x}\right)}{{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-\left(1+\sin 2\mathrm{x}\right)-\left(1-\sin 2\mathrm{x}\right)}{{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}^2} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{-1-\sin 2\mathrm{x}-1+\sin 2\mathrm{x}}{{\left(\cos \mathrm{x}+\sin \mathrm{x}\right)}^2} \\ \Rightarrow \boxed{\frac{\mathbf{dy}}{\mathbf{dx}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\frac{\mathbf{2}}{{\left(\mathbf{cos}\mathbf{}\mathbf{x}\mathbf{}\mathbf{+}\mathbf{}\mathbf{sin}\mathbf{}\mathbf{x}\right)}^{\mathbf{2}}}} \end{gathered}$$