Question

What is the derivative of \[\arctan \left( {\dfrac{1}{x}} \right)\] ?

Answer 1

Hint: Here the arc function is the inverse function. Also we know the value of tan inverse function. Only the function on which the tan inverse is operated is not x as it is reciprocal of x we can say. But we will use the same formula as we use for the derivative of tan inverse.
Formula used:
 \[\dfrac{d}{{dx}}{\tan ^{ - 1}}\left( x \right) = \dfrac{1}{{1 + {x^2}}}\]

Complete step by step solution:
Given the function is,
 \[\arctan \left( {\dfrac{1}{x}} \right)\]
This is nothing but,
 \[\arctan \left( {\dfrac{1}{x}} \right) = {\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)\]
We have to find the derivative of the function so given that is
 \[ = \dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)\]
We know that, \[\dfrac{d}{{dx}}{\tan ^{ - 1}}\left( x \right) = \dfrac{1}{{1 + {x^2}}}\]
So here we can write \[x = \dfrac{1}{x}\] ,
 \[ = \dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right)\]
 \[ = \dfrac{1}{{1 + {{\left( {\dfrac{1}{x}} \right)}^2}}}\]
Now taking the square,
 \[ = \dfrac{1}{{1 + \dfrac{1}{{{x^2}}}}}\]
Taking the LCM we can write,
 \[ = \dfrac{1}{{\dfrac{{{x^2} + 1}}{{{x^2}}}}}\]
Now the denominator of the fraction will shift to the numerator,
 \[ = \dfrac{{{x^2}}}{{1 + {x^2}}}\]
So we get the correct answer,
So, \[\dfrac{d}{{dx}}{\tan ^{ - 1}}\left( {\dfrac{1}{x}} \right) = \dfrac{{{x^2}}}{{1 + {x^2}}}\]
So, the correct answer is “ \[\dfrac{{{x^2}}}{{1 + {x^2}}}\] ”.

Note: Note that the function so given is the regular function only the tan operated function is changed. Also note that all the trigonometric functions have their own inverse function and their respective derivatives also. Whenever the function is other than x ; make changes as we have done above.