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

## Show that the function $y = \frac{1}{1 + \tan x}$ is decreasing for all values of $x$? Thanks!

##### 1 Answer
Oct 3, 2017

Start by finding the derivative.

$y = {\left(\tan x + 1\right)}^{-} 1$

By the chain rule, we have

$y ' = - {\sec}^{2} \frac{x}{\tan x + 1} ^ 2$

We immediately see that ${\sec}^{2} x$ will always be positive no matter the value of $x$. Furthermore, ${\left(\tan x + 1\right)}^{2}$ 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 $x \ne \frac{3 \pi}{4} \mathmr{and} \frac{7 \pi}{4}$.

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

Hopefully this helps!