Question

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

Answer 1

`y=sqrt(1+x^2)=(1+x^2)^(1/2)`

Use the chain rule to differentiate:

`dy/dx=1/2(1+x^2)^(-1/2)d/dx(1+x^2)=x(1+x^2)^(-1/2)=x/sqrt(1+x^2)`

Using the product and chain rules:

`(d^2y)/dx^2=(1+x^2)^(-1/2)-1/2x(1+x^2)^(-3/2)(2x)`

`color(white)((d^2y)/dx^2)=(1+x^2)^(-1/2)-x^2(1+x^2)^(-3/2)`

`color(white)((d^2y)/dx^2)=1/(1+x^2)^(1/2)-x^2/(1+x^2)^(3/2)`

`color(white)((d^2y)/dx^2)=(1+x^2-x^2)/(1+x^2)^(3/2)`

`color(white)((d^2y)/dx^2)=1/(1+x^2)^(3/2)`

The domain of `y` is `x in RR`.

We see that `dy/dx=0` at `x=0`. When `x<0`, `dy/dx<0` and when `x>0`, `dy/dx>0`.

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

`(d^2y)/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_(xrarr-oo)y=lim_(xrarr+oo)y=+oo`.

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 `(0,1)`, and switches from increasing to decreasing at `x=0`.