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

Question

33539 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:

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

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