Question

if y = tanx+cotx/tanx-cotx then dy/dx

Answer 1

Dear student,

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

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