What is the derivative of arctan(x)+arctan(1/x)?

1 Answer
mason m
Jun 11, 2016

#0#

Explanation:

Alternatively, we can simplify the original function.

#y=arctan(x)+arctan(1/x)#

Take the tangent of both sides.

#tan(y)=tan(arctan(x)+arctan(1/x))#

Use the tangent addition formula: #tan(alpha+beta)=(tan(alpha)+tan(beta))/(1-tan(alpha)tan(beta))#

Here, for #tan(arctan(x)+arctan(1/x))#, we see that #alpha=arctan(x)# and #beta=arctan(1/x)#, so we obtain:

#tan(y)=(tan(arctan(x))+tan(arctan(1/x)))/(1-tan(arctan(x))tan(arctan(1/x))#

#tan(y)=(x+1/x)/(1-x(1/x))#

#tan(y)=((x^2+1)/x)/(1-1)#

#tan(y)=(x^2+1)/0#

This is an undefined value: however, we know that the tangent of #y# is undefined, and the tangent function is undefined at #pi/2# and #-pi/2#.

So, we know that

#y=pi/2" "# or #" "y=-pi/2#

Thus,

#arctan(x)+arctan(1/x)=+-pi/2#

And the derivative of #pi/2# or #-pi/2#, which are both constant, is just #0#.

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