How do you find the second derivative of #y=tan(x)#?

Question

33157 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-04T03:46:16

#2sec^2xtanx#

First we find #d/dxtanx#.

We know that #tanx=sinx/cosx#

So we can use the quotient rule to solve for this:

#d/dx(sinx/cosx)=(cosxd/dx(sinx)-sinxd/dx(cosx))/cos^2x#

#color(white)(d/dx(sinx/cosx))=(cosx(cosx)-sinx(-sinx))/cos^2x#

#color(white)(d/dx(sinx/cosx))=(cos^2x+sin^2x)/(cos^2x)=1/cos^2x#

#d/dx(sinx/cosx)=d/dxtanx=sec^2x#

Now for #d^2/dx^2tanx#, or #d/dxsec^2x#

Which we can write as #d/dx(secx)^2#, which gives:

#2secx(secxtanx)#, using the chain rule, where we compute #d/(du)u^2# and #d/dxsecx#.

Which gives:

#2sec^2xtanx#

So:

#d^2/dx^2tanx=2sec^2xtanx#