Question

how do I solve g(x)=5cos^2((pi)t) ??

Answer 1

derivative

Answer 2

huh?

Answer 3

im sorry but I don't understand u, loser66...

Answer 4

yes sorry, sorry, sorry :(

Answer 5

its g(t)=5cos^2((pi)t)

Answer 6

kinda...

Answer 7

u can show me ur way, itll probably be helpful...

Answer 8

ok, I think I see it...

Answer 9

hold on a lil im trying to understand u a lil more

Answer 10

what happened to the 5?

Answer 11

lol its ok, but im actually really lost :((

Answer 12

2

Answer 13

sorry I forgot the x, I get that..

Answer 14

3x

Answer 15

3x^2

Answer 16

sorry

Answer 17

7x^6

Answer 18

sorry and its 2cos ??

Answer 19

:)

Answer 20

-sin

Answer 21

so -2sin?

Answer 22

so -2cos(t)sin(t) ??

Answer 23

2 ?

Answer 24

2cos ??

Answer 25

oh, so -2sin(t) ?

Answer 26

ok I get that :)

Answer 27

ummm... -4sin(t) ?

Answer 28

ok :(

Answer 29

\[g(t)=5\cos^2(\pi(t))\]

Answer 30

wait a minute,

Answer 31

ok

Answer 32

I am so sorry. i think this way is easier than chain rule, but it seems I confuse you. If you don't get it, let tag other one. @zepdrix

Answer 33

what, what's going on? :U

Answer 34

I tried to help but my dummy way confuse the asker, help me friend

Answer 35

sorry :(

Answer 36

\[\Large g(x)=5\cos^2\left(\pi t\right)\]Find g'(x)? :O hmm

Answer 37

no sorry its g(t)

Answer 38

Ah yes :) my mistake

Answer 39

sorry, wrote it wrong :/

Answer 40

So we have to apply the `chain rule` several times.
Do you understand that the `outermost function` is the ( )^2?
If not, we can write our function like this,\[\Large g(t)=5\left[\cos(\pi t)\right]^2\]

Answer 41

ok, can we do it like that?

Answer 42

That power placement on trig functions can be a little tricky :U
So when we take our derivative, we start by applying the `power rule` to the `outermost function`.\[\Large g'(t)=5\cdot2\left[\cos(\pi t)\right]\]But the chain rule tells us to multiply by the derivative of the inner function,\[\Large g'(t)=5\cdot2\left[\cos(\pi t)\right]\color{royalblue}{\left[\cos(\pi t)\right]'}\]

Answer 43

thank you though @Loserr66

Answer 44

@loser66 ^

Answer 45

Winner66 \:D/

Answer 46

watever,

Answer 47

:/

Answer 48

So we need to take the derivative of that blue portion still. :O
Understand the process kinda?
It takes a little getting used to.

Answer 49

@zepdrix still have (pi t)'

Answer 50

Yah we haven't taken the derivative of the cosine part yet :3 (pi t)' coming soon :D

Answer 51

derivative of cos(pi t)= -sin(pi) ??

Answer 52

Whenever we take the derivative of our `outer function`, we leave the inside alone.
So in this case, cosine is our outer function.

Differentiating cosine gives us -sine, yes.
But the inside should stay the same.

Answer 53

\[\Large g'(t)=5\cdot2\left[\cos(\pi t)\right]\color{orangered}{\left[-\sin(\pi t)\right]}\]Ok good! But we have to apply the chain rule again! :O\[\Large g'(t)=5\cdot2\left[\cos(\pi t)\right]\color{orangered}{\left[-\sin(\pi t)\right]}\color{royalblue}{\left(\pi t\right)'}\]

Answer 54

si its, 10(cos(pi t))(-sin(pi t))(pi) ??

Answer 55

Yes, good.
Now since multiplication is commutative, we can rearrange things so it looks a little nicer.
Bring the pi and negative sign to the front.

Answer 56

\[-10\pi(\cos(\pi t))(-\sin(\pi t))\]

Answer 57

?? yes, orr no?:O

Answer 58

Woops, when you moved the negative sign to the front, you accidentally left it on the sine also lol.
\[\large g'(t)=-10\pi \cos(\pi t)\sin(\pi t)\]

Answer 59

oh true, sorry about that :o

Answer 60

yay team \c:/

Answer 61

and that's it?? :O -_____-

Answer 62

thaaaatttt 's iiiiiit??? nope, I am struggling with it, not just "that'/s/it" friend

Answer 63

lol what??

Answer 64

ok, I am mean person!! sorry, I take it back, yeah, you are right, "That's it" thanks @zepdrix

Answer 65

:x

Answer 66

so @zepdrix dats da final answer?

Answer 67

ya :o

Answer 68

I feel sooo stupid!!! :(

Answer 69

ya dat wud da final answer hahahaha.... how fun it is, teen words, right?

Answer 70

haha sorry :p just used to them...

Answer 71

thank you both! @zepdrix @loser66 :)

Answer 72

water you talkking about loser? o3o
doesn't everione talk like dat?