# How do you sketch the graph y=sqrt(1+x^2) using the first and second derivatives?

Aug 24, 2017

$y = \sqrt{1 + {x}^{2}} = {\left(1 + {x}^{2}\right)}^{\frac{1}{2}}$

Use the chain rule to differentiate:

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

Using the product and chain rules:

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 = {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} - \frac{1}{2} x {\left(1 + {x}^{2}\right)}^{- \frac{3}{2}} \left(2 x\right)$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{\mathrm{dx}} ^ 2} = {\left(1 + {x}^{2}\right)}^{- \frac{1}{2}} - {x}^{2} {\left(1 + {x}^{2}\right)}^{- \frac{3}{2}}$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{\mathrm{dx}} ^ 2} = \frac{1}{1 + {x}^{2}} ^ \left(\frac{1}{2}\right) - {x}^{2} / {\left(1 + {x}^{2}\right)}^{\frac{3}{2}}$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{\mathrm{dx}} ^ 2} = \frac{1 + {x}^{2} - {x}^{2}}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

$\textcolor{w h i t e}{\frac{{d}^{2} y}{\mathrm{dx}} ^ 2} = \frac{1}{1 + {x}^{2}} ^ \left(\frac{3}{2}\right)$

The domain of $y$ is $x \in \mathbb{R}$.

We see that $\frac{\mathrm{dy}}{\mathrm{dx}} = 0$ at $x = 0$. When $x < 0$, $\frac{\mathrm{dy}}{\mathrm{dx}} < 0$ and when $x > 0$, $\frac{\mathrm{dy}}{\mathrm{dx}} > 0$.

The function is decreasing on $x < 0$, levels out at $x = 0$, and is increasing for $x > 0$.

$\frac{{d}^{2} y}{\mathrm{dx}} ^ 2 > 0$ for all Real values of $x$, so the function is always concave up.

We can also note that $y$ itself has no Real zeros. For all Real values of $x$, $y > 0$. The minimum value will then be at $x = 0$, where $y = 1$. Note that ${\lim}_{x \rightarrow - \infty} y = {\lim}_{x \rightarrow + \infty} y = + \infty$.

The drawing of the graph should then look something like:

graph{sqrt(1+x^2) [-17.34, 18.7, -4.97, 13.05]}

The graph is always concave up, has a minimum at $\left(0 , 1\right)$, and switches from increasing to decreasing at $x = 0$.