Question

Please help, i need to find the derivative of y=1/(1+sec(x))^2.

Answer 1

\[\frac{ d }{ dx }\frac{ 1 }{ (1+\sec(x))^2 }\] this?

Answer 2

yes, essentially

Answer 3

You can bring it up and use chain rule

Answer 4

power + chain

Answer 5

Give it an attempt and see what you get, then we'll see if you did it right or not.

Answer 6

i know the chain rule, how do i bring it up?

Answer 7

Exponent rules, if you remember \[\frac{ 1 }{ x^n } \implies x^{-n}\]

Answer 8

omg... haha yes, ok, ill give it a show

Answer 9

:)

Answer 10

Yeah I think it's easiest if you bring it up in the numerator

Answer 11

is it y'=tan(x)/4?

Answer 12

Exponent rules, if you remember \[\left[ (1+\sec(x))^{-2} \right]' \implies -2(1+\sec(x))^{-3} \times (1+\sec(x))'\] did you get this far?

Answer 13

i made an error in the exponent rule, one second

Answer 14

Np, take your time

Answer 15

Oh Idk why, but the lag removed my information above, w/e so you also have to keep in mind

Answer 16

\[\frac{ d }{ dx } \sec(x) = \sec(x)\tan(x)\]

Answer 17

my notes say \[d/dx(\sec(g(x))) = \sec(g(x))\tan(g(x))*g'(x)\]
now i have \[y'=\frac{ \sec(x)\tan(x) }{ -2(1+\sec(x))^3 }\]

Answer 18

is there anything more to simplify? if im at the right spot

Answer 19

-2 should be in the numerator but other than that, you're good :)

Answer 20

ahh ok, because its not to the -3rd power

Answer 21

Right!

Answer 22

Derivatives aren't so bad, it's just bunch of rules you remember and then just algebra

Answer 23

thank you so much, honestly. and i know that, its exactly the fact that i cant remember a few of the rules, and just get blocked

Answer 24

No worries :)

Answer 25

A lot of people forget exponent rules especially so I'm going to put some up right now :P

Answer 26

Glad my mistakes could help a few more :)