Question

I don't understand how f(x)=(tan(x)-2)/sec(x) gives f'(x)= 2sin(x)cos(x). Can someone show me how this was done. Please and thank you.

Answer 1

Do you know your tangent and secant derivatives?

Answer 2

tangent is sec^2x and secant is tanxsecx

Answer 3

(f'g-g'f)/g^2 so (sec^2x(secx)-tanxsecx(tanx))/sec^2x
right?

Answer 4

Oh oh oh I see the problem.

Answer 5

Yah there was a little boo boo in your derivative.

Answer 6

\[\Large\rm =\frac{\color{royalblue}{\left(\tan x-2\right)'}\sec x-(\tan x-2)\color{royalblue}{\left(\sec x\right)'}}{\sec^2x}\]
\[\Large\rm =\frac{\color{orangered}{\left(sec^2x\right)}\sec x-(\tan x-2)\color{orangered}{\left(\sec x tan x\right)}}{\sec^2x}\]Do you see how this is a little different than what you wrote?

Answer 7

You forgot the -2 from the tanx-2 in the second term.

Answer 8

Your right. Oups

Answer 9

Still factoring out that secant is a good idea,
and cancelling out if able,\[\Large\rm =\frac{\cancel{\sec x}\left(\sec^2x-(\tan x-2)\tan x\right)}{\sec^{\cancel{2}1}x}\]

Answer 10

Which leaves us with this:
\[\Large\rm =\frac{\color{royalblue}{\sec^2x-\tan^2x}+2\tan x}{\sec x}\]

Answer 11

And we used our Pythagorean Identity to show that this blue part is simply 1,\[\Large\rm =\frac{\color{royalblue}{1}+2\tan x}{\sec x}\]

Answer 12

Mmmm what do you think? :d
Able to simplify it the rest of the way from there?
Confused by any of those steps?

Answer 13

no I am following. Makes more sense now that the 2 is there.

Answer 14

Thank you very much for your help

Answer 15

No problem!