Mathematicsy=cot-1 {tan (π/2-x)} then dydx =

$y = \cot^{- 1} {\tan (π / 2 - x)}$ then $\frac{dy}{dx} =$

A
x
B
1
C
$\frac{1}{1 + x^{2}}$
D
$\frac{- 1}{1 + x^{2}}$

Solution:

The given function is $y = c o t^{- 1} (\tan (\frac{π}{2} - x))$

This can be rewrite as $\begin{array}{rcl} y & = & c o t^{- 1} (c o t x) = x \end{array}$

Differentiate both sides with respect to $x$

$\frac{d y}{d x} = 1$

Related content

Matrices and Determinants_mathematics

Critical Points Solved Examples

Type of relations_mathematics