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

Question

Paisley Johnson · Accepted Answer

#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#.