F(x)=2secx+tanx 0

find the derivative?

set it =0

i know that... how do i find the derivatives.

eyust707 take the derivative for her :0)

okay

you find the derivative of each term seprate

lets look at the first term

secx=secxtanx and tanx=secx^2. but what do you do with the 2 in front of the secx

the two is a constant is just hangs out in front

yup

so, when finding derivatives it just stays? so the answer for the derivative would be 2secxtanx + secx^2?

yep

you don't have to use the product rule or whatever?

product rule is when you there are two variable in the same term

and 2 isn't a variable? ha. alright. got it.

like tanxsecx would need product rule tanx+secx would not

yeah i know. but the 2 is being multiplied...

yea but the 2 is not a variable.. its a number

but the derivative of a number is 0. that's what messes me up

ahhh i see i see

so it just hangs out there then. well that's dumb. ha

yeah just like the derivative of 5*x^2

is 10x

oh got it. nice example.

the five stays bring the two down you get 2*5x

yep!

so when you are finding the critical points...

yes..

you set it equal to zero... so it would be 0=2secxtanx+secx^2. bring the secx out?

secx(2tanx+secx)?

yea its been a little wile since i did that but your math is def correct

good job brie

I just don't get it. then you would have to set the inside of the parenthesis to zero... and i don't know how to do that.

because they are two different trig terms.

2tanx+secx=0

yeah. and then what

well lets see not 100 percent sure but..

2tanx=-secx

tanx=sinx/cosx and secx = 1/cosx

brilliant! alright. thank you

wait wait wait. still don't get it.

2sinx/cosx+1/cosx=2sinx/cosx

its been hella days since i did these alvin you got these down?

eliminate one of the trig functions and leave x on one side i think

yeah... ha. but which one can be eliminated?

sinx+1/cosx

2tanx+cosx=1

2tanx=1-cosx

i thought it equaled 0?

i think theres a rule for 1-cosx??

hehe i messed up

2tanx+secx=0

thought you took the derivative before?

that is what we ended up with

now we are trying to find the critical points.

for the derivative we ended up with secx(2tanx+secx)=0

lets start from where we were 2tanx=-secx 2tanx= -1/cosx 2tanxcosx= -1 tanxcosx=-1/2 sinxcosx/cosx=-1/2

brilliant!

sooo...

so sinx=-1/2 and that's at 7pii/6

and 11pii/6

for the secx=0

1/cosx=0?

that DNE

why?

1 over what number gives you zero?

because 1 doesn't equal 0 when you multiply both sides by cosx?

is there any number you can put in?

right.. got it. thank you.

well infinity would work... butt cosx is never greater than 1

kk i got a midterm in the mornin i gots to study for but if you got skype shoot me a message on there and id be glad to help. my name is exactly the same on there

wait... did we mess up

lol maybe

ohh nvm haha. thank gosh

thanks for all the help.

welcome =)

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