What is the derivative of #f(x)=cot^-1(x)# ?

Answer 1

By Implicit Differentiation,

#f'(x)=-1/{1+x^2}#

Let us look at some details.

By replacing #f(x)# by #y#,
#y=cot^{-1}x#

by rewriting in terms of cotangent,

#Rightarrow coty=x#

by implicitly differentiating with respect to x,

#Rightarrow -csc^2ycdot{dy}/{dx}=1#
by dividing by #-csc^2y#,
#Rightarrow {dy}/{dx}=-1/{csc^2y}#
by the trig identity #csc^2y=1+cot^2y=1+x^2#,
#Rightarrow {dy}/{dx}=-1/{1+x^2}#

Hence,

#f'(x)=-1/{1+x^2}#
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Answer 2

The derivative of ( f(x) = \cot^{-1}(x) ) is ( -\frac{1}{x^2 + 1} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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