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What is the derivative of #f(x)=arctan[(x+1)/(x-1)]#?

Answer 1

Emma Aldridge

# rArr f'(x)=-1/(x^2+1), (x!=1)#.

Observe that,

#tan(pi/4+theta)=(tan(pi/4)+tantheta)/(1-tan(pi/4)tantheta)=(1+tantheta)/(1-tantheta)#.

#:. tan(-(pi/4+theta))=-tan(pi/4+theta)=(tantheta+1)/(tantheta-1)#.

Recall that, the range of #tan# function is #RR#.

So, we can very well take #x=tantheta# and get,

#f(x)=arc tan{(x+1)/(x-1)}=arc tan{(tantheta+1)/(tantheta-1)}#,

#=arc tan{tan(-(pi/4+theta)}#,

#=-(pi/4+theta)#.

# rArr f(x)=-pi/4-arc tanx, (x!=1)#.

# rArr f'(x)=0-1/(x^2+1)=-1/(x^2+1), (x!=1)#, is the

desired derivative!

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Answer 2

Samantha Adkison

The derivative of ( f(x) = \arctan\left[\frac{x+1}{x-1}\right] ) is ( f'(x) = \frac{1}{1 + \left(\frac{x+1}{x-1}\right)^2} \cdot \frac{(x-1) - (x+1)}{(x-1)^2} ).

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Answer 3

Scarlett Allison

The derivative of ( f(x) = \arctan\left(\frac{x+1}{x-1}\right) ) is ( \frac{1}{1 + \left(\frac{x+1}{x-1}\right)^2} \cdot \frac{(x-1) - (x+1)}{(x-1)^2} ).

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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.