Show that the function #y=1/(1+tanx)# is decreasing for all values of #x#?

Show that the function #y=1/(1+tanx)# is decreasing for all values of #x#?

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1 Answer
Noah G
Oct 3, 2017

Start by finding the derivative.

#y = (tanx + 1)^-1#

By the chain rule, we have

#y' = -sec^2x/(tanx + 1)^2#

We immediately see that #sec^2x# will always be positive no matter the value of #x#. Furthermore, #(tanx + 1)^2# will be positive no matter what value of #x#.

This means that #y'# will be negative wherever the function/derivative is defined, hence the function is uniformly decreasing.

However, with restrictions we see that #x!= (3pi)/4 or (7pi)/4#.

So the function is decreasing but then there are asymptotes, so just keep that in mind. Here is the graph to verify.

enter image source here

Hopefully this helps!

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