Question

How do you find the derivative of \[y = \operatorname{arccot} \left( x \right)\] ?

Answer 1

Hint:$\dfrac{d}{{dx}}\left[ {\operatorname{arccot} \left( x \right)} \right] = - \dfrac{1}{{1 + {x^2}}}$

Notice if \[y = \operatorname{arccot} \left( x \right)\] then,
$\sin (y) = \dfrac{1}{{\sqrt {1 + {x^2}} }}$
Using Pythagoras Theorem
${\left( {hypotenuse} \right)^2} = {\left( {base} \right)^2} + {\left( {height} \right)^2}$
we get, $\cos (y) = \dfrac{x}{{\sqrt {1 + {x^2}} }}$
Complete step by step answer-
 Given that \[y = \operatorname{arccot} \left( x \right)\], then it follows
$\cot (y) = x$ and $0 < y < \pi $
Differentiating both sides with respect to x we get
$ \Rightarrow \dfrac{d}{{dx}}\left[ {\cot \left( y \right)} \right] = \dfrac{d}{{dx}}\left[ x \right]$
$ \Rightarrow - \cos e{c^2}\left( y \right).\dfrac{{dy}}{{dx}} = 1$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{{\cos e{c^2}\left( y \right)}} = - {\sin ^2}\left( y \right)$
$ \Rightarrow \dfrac{{dy}}{{dx}} = - {\left( {\dfrac{1}{{\sqrt {1 + {x^2}} }}} \right)^2} = - \dfrac{1}{{1 + {x^2}}}$
Therefore, we get $\dfrac{d}{{dx}}\left[ {\operatorname{arccot} \left( x \right)} \right] = - \dfrac{1}{{1 + {x^2}}}$
Note- It is advised to the students to learn the value of the derivative of \[y = \operatorname{arccot} \left( x \right)\] as it will be useful in the future.