Question

Given F(2)=1, F'(2)=9, F(4)=3, F'(4)=7 and G(4)=2, G'(4)=6, G(3)=4, and G'(3)=9, find....

Part 1: H(4) if H(x)=F(G(x))

Part 2: H ' (4) if H(x)=F(G(x))

Part 3: H(4) if H(x)=G(F(x))

Part 4: H ' (4) if H(x)=G(F(x))

Part 5: H ' (4) if H(x)= F(x)
-------
G(x)

please explain!! :)
thank you! :D

Answer 1

Wasn't going to log in, but then I saw you online :>
I don't really see why those derivatives are necessary, but if
H(x) = F(G(x))
Then shouldn't H(4) simply be F(G(4))?
LOL

Answer 2

because thats the 1st part of a 5 part question
derivatives will be useful in other parts

Answer 3

Yeah, didn't see that part until the last minute lol

my bad.

But what say you, food lover? :3

Answer 4

post all of them at once

Answer 5

hahaa thanks for logging on :)

and yeah what @hartnn said :P

ohhh so would i be solving for F(2) since G(4)=2?

Answer 6

Good so far, the answer is then....?

Answer 7

ermm.. F(2)=1

so H(4)=1 ?

and i've edited it with all five parts:)

Answer 8

thats correct, H(4) is indeed 1
for 2nd part, you'll need chain rue, know what that is ?

Answer 9

awesome!! :)

it's where you work from the outer function first right?

but i'm not too good at it haha :(

Answer 10

chain rue is a combination of different rues LUL

Answer 11

have you come up with a technique how to approach this problem?

Answer 12

not really sure how to start it off :(

Answer 13

Chain rule :
\[\Large \frac{d}{dx}f\left(g(x)\right)=f'\left(g(x)\right) g'(x)\]

Answer 14

will tell you how to do part 2,
you try part 3,4 by yourself.

chain rule written with other notation:
\([F(G(x))]' = F'(G(x)) \times G'(x)\)
i think you can find the value of terms on right side, try it

Answer 15

ohhh okay

yeah, I'm not too good with the chain rule... yet :P

so F'(G(4)) x G'(4) = F'(2) x 6 = 9 x 6 = 54 ?

Answer 16

okay @hartnn :)

so would part 3 be this?

H(4) if H(x)=G(F(x))

H(4)=G(F(4))
=G(3)
=4

So for part 3, H(4)=4 ?

does that look right?

Answer 17

Ohh you.

Answer 18

hahah :P

Answer 19

wait so did I not do part 2 right? :/

Answer 20

yes, both 2nd , 3rd part are correct :)

Answer 21

oh yay!! :)

lemme try part 4.. just a moment :)

Answer 22

If I knew hart was going to butt in, I wouldn't have logged on XD

LOL JK @hartnn ^_^

Signing off now, anyway, good luck, food lover, and TJ out :)

Answer 23

lawlzz okay, thanks @terenzreignz :)

see you later!!

Answer 24

did i just scare him away :O :P :(

Answer 25

someone should do a tutorial lesson on calculus

Answer 26

so would i start part 4 like this?

H ' (4) if H(x)=G(F(x))

[G(F(x))]' = G'F(x) * F ' (x) ?

and @hartnn i don't think so hahaha :P

and @nincompoop, yes! though i would definitely not nominate myself for that job... for obvious reasons lol

Answer 27

yep, G'F(x) * F ' (x)
G'(F(x)) * F ' (x) , careful with brackets, else you might get confused later or could get a wrong answer..

Answer 28

ohh okay will keep that in mind :)

so now it looks like this?

G'(F(4)) * F ' (4)
= G'(3) * 7
=9 * 7 =63

so H ' (4) for part 4 equals 63?

Answer 29

sure it is :)

Answer 30

yay!! :)

so last part now!! how would this one work? :/

Answer 31

know quotient rule ?

Answer 32

\[\frac{ u'v-uv' }{ v^2 }\] ?

Answer 33

yes :) use it

Answer 34

okay:)

so would f(x) = u and g(x)=v ?

Answer 35

go on, i will stop u , if u make an error...

Answer 36

okay:)

and then

(f ' (4) * g(4)) - (f(4) * (g ' (4))
-------------------------
g(4)^2

?

Answer 37

and then

(7 * 2) - (3 * 6)
-------------
2^2

= 14 - 18
---------
4

= -4/4 = -1?

so H'(4) for H(x)= f(x)/g(x) for part 5 equals -1?

did i do that right?

Answer 38

seems correct to me :)

Answer 39

ahh yay!! and we're done!! thanks so much!!!! :D

Answer 40

welcome ^_^