# Question #c66f7

Oct 13, 2016

Decreasing on $\left(- \infty , 0\right)$ and increasing on $\left(0 , \infty\right)$.

#### Explanation:

The derivative of a function tells us about whether that function is increasing or decreasing.

• If $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$, then $y$ is increasing.
• If $\frac{\mathrm{dy}}{\mathrm{dx}} < 0$, then $y$ is decreasing.

So, we need to take the derivative of the function and find when it is positive and when it is negative.

To find the derivative of the given function, we will need the chain rule. In the case of a natural logarithm function, as we have here, we see that:

$\frac{d}{\mathrm{dx}} \ln \left(x\right) = \frac{1}{x} \text{ "=>" } \frac{d}{\mathrm{dx}} \ln \left(u\right) = \frac{1}{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

So, we have:

$y = \ln \left(1 + {x}^{2}\right) \text{ "=>" } \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{1 + {x}^{2}} \cdot \frac{d}{\mathrm{dx}} \left(1 + {x}^{2}\right) = \frac{2 x}{1 + {x}^{2}}$

So, we need to find when $\frac{2 x}{1 + {x}^{2}}$ is positive and when it is negative. We can find the boundaries of these positive and negative intervals by first finding when the function equals $0$ or becomes undefined.

$\frac{2 x}{1 + {x}^{2}} = 0 \text{ "=>" "2x=0" "=>" } x = 0$

This, $x = 0$, is one endpoint to our intervals. The function is undefined when $1 + {x}^{2} = 0$, which never occurs for any real value of $x$.

So, all we have to do is examine $\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{2 x}{1 + {x}^{2}}$ around the point $x = 0$.

When $x < 0$, for example, at $x = - 1$, we see that the derivative equals:

$\frac{\mathrm{dy}}{\mathrm{dx}} {|}_{x = - 1}$ $= \frac{2 \left(- 1\right)}{1 + {\left(- 1\right)}^{2}} = \frac{- 2}{2} = - 1$

Since this is negative, we know that $\frac{\mathrm{dy}}{\mathrm{dx}} < 0$ on the entire interval of $\left(- \infty , 0\right)$. This means that $y$ is decreasing on this interval.

Similarly, checking $\frac{\mathrm{dy}}{\mathrm{dx}}$ when $x > 0$, we see that $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$ and thus $y$ is increasing on the interval $\left(0 , \infty\right)$.