Question

y = (tan x +cot x)/ (tan x - cotx ) then dy/dx =?

Answer 1

$\text{First we simplify the given function, it will make the differentiation very simple}$

$\textbf{Given:}$

$y=\displaystyle\frac{tanx+cotx}{tanx-cotx}$

$y=\displaystyle\frac{tan\,x+\frac{1}{tan\,x}}{tan\,x-\frac{1}{tan\,x}}$

$y=\displaystyle\frac{\frac{1+tan^2x}{tan\,x}}{\frac{tan^2x-1}{tan\,x}}$

$y=\displaystyle\frac{1+tan^2x}{tan^2x-1}$

$y=\displaystyle\frac{1+tan^2x}{-(1-tan^2x)}$

$y=\displaystyle\frac{1}{-\frac{(1-tan^2x)}{1+tan^2x}}$

$y=\displaystyle\frac{1}{-cos\,2x}$

$y=\displaystyle\,-sec\,2x$

$\text{Differentiate with respect to x}$

$\displaystyle\frac{dy}{dx}=-2\,sec\,2x\,tan\,2x$