Question

LGBADERIVATIVE:

\[\huge e^{\theta} (\tan \theta - \theta)\]

i have no clue..maybe product rule?

Answer 1

yeah, product rule

Answer 2

\[\huge e^\theta \tan \theta = e^\theta \theta\]
then twice product rule?

Answer 3

that equal is a minus sign of course

Answer 4

no need to distribute

Answer 5

no?

Answer 6

oh i see

Answer 7

\[u=e^{\theta}\]\[v=(\tan\theta-\theta)\]

Answer 8

u = \(e^\theta\)
v = \((\tan \theta - \theta)\)

Answer 9

yeah what i meant

Answer 10

ditto!

Answer 11

so it's \[e^\theta (\sec^2 \theta + 1) + e^\theta (\tan \theta -1)?\]

Answer 12

why the -1 in the second set of parentheses?

Answer 13

or the +1 in the first...?

Answer 14

oh it's supposed to be theta

Answer 15

and that +1 should be -1

Answer 16

\[e^\theta \tan \theta + e^\theta (\tan \theta - \theta)\]

Answer 17

is that better?

Answer 18

\[u=e^x\implies u'=e^x\]\[v=\tan x-x\implies v'=\sec^2x-1\]\[(uv)'=u'v+v'u=e^x(\tan x-x)+e^x(\sec^2x-1)\]

Answer 19

so im right right? since sec^2 - 1 = tan^2

Answer 20

oh darn another typo >.<

Answer 21

darn latex

Answer 22

oh yeah, lol

Answer 23

now I see what you did

Answer 24

It's ok I make those too @lgbasallote , Turing is just a living android so he doesn't make syntax errors :-3

Answer 25

so just to be clear
\[e^\theta (\tan^2 \theta) + e^\theta (\tan \theta - \theta)?\]

Answer 26

agree @agentx5 darn androids

Answer 27

@agentx5 yeah right... I have everything copy pasted to erase my errors quickly is all ;)

Answer 28

yes @lgbasallote

Answer 29

:-D

But as long as it's "first d second + second d first" you're good

Answer 30

thanks! :D

Answer 31

Oh here's one more for you @lgbasallote :
Quotient rule (how I remember it)

Answer 32

Quotient rule:
\[\frac{d}{dt}\frac{high}{low} = \frac{((low)d(high)-(high)d(low)}{(low)^2} \]

Answer 33

It kind of has a sing-song tone to it, makes it easy to remember

Answer 34

I like to derive the quotient rule, I almost never even use it
use the product rule instead

Answer 35

Eeek! Yeah I try to avoid making things unnecessarily harder :-D But to each their own technique that works for them, and yours clearly works for you

Answer 36

let \[\frac uv=(uv^{-1})\]\[(uv^{-1})'=u'v^{-1}-v^{-2}v'u={u'v-v'u\over v^2}\]i.e. product rule+chain rule=quotient rule.
but whatever is easiest for you is always best :)