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Mathematics 18 Online
OpenStudy (anonymous):

Strange question.. Take the derivative y=(x=2)^8 (x+3)^6

hartnn (hartnn):

product rule and x^n rule

OpenStudy (anonymous):

I wonder if the prof. ended up mistyping. No? really?

OpenStudy (anonymous):

wow, I've never seen something like this

hartnn (hartnn):

is that x=2 inside the brackets ?

hartnn (hartnn):

i think thats \(\Large y = (x+2)^8(x+3)^6\)

OpenStudy (anonymous):

see, the thing is that it really is an equal sign :/

OpenStudy (anonymous):

I think it was a typing error

hartnn (hartnn):

its definitely their typing mistake

OpenStudy (anonymous):

good haha XD

hartnn (hartnn):

now can you solve it ?

OpenStudy (anonymous):

chain rule?

hartnn (hartnn):

product rule + chain rule + x^n rule

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

ok so

hartnn (hartnn):

if you know those, this is really simple :)

Miracrown (miracrown):

ok so for this one we definitely need the chain rule and the product rule like hartnn stated.

OpenStudy (anonymous):

1st step is

hartnn (hartnn):

\(\Large (fg)' - f'g +fg'\) here f = (x+2)^8 g = (x+3)^6

hartnn (hartnn):

\(\Large (fg)' = f'g +fg'\)

OpenStudy (anonymous):

8(x+2)^7 (x+3)^6 + (x+2)^8 6(x+3)^5

hartnn (hartnn):

yes! :)

OpenStudy (anonymous):

this?

OpenStudy (anonymous):

ok cool

Miracrown (miracrown):

well the product rule says to take the derivative of one part and times it by the other part and then add that to the derivative of the other part times the first part.

hartnn (hartnn):

now factor out terms, whichever you can

OpenStudy (anonymous):

well the (x+2) and (x+3)s

hartnn (hartnn):

and 2

OpenStudy (anonymous):

ah yes

Miracrown (miracrown):

yessss xD

OpenStudy (akashdeepdeb):

Use logarithms. It becomes much easier. :D

OpenStudy (anonymous):

lol no u

hartnn (hartnn):

only if you love logs :P

OpenStudy (anonymous):

hmm so then it is

Miracrown (miracrown):

not for taking the derivative

hartnn (hartnn):

logs can be used here, but prolly not the best method..

OpenStudy (akashdeepdeb):

Oh, okay. :)

Miracrown (miracrown):

yeahhh

OpenStudy (anonymous):

2(x+2)(x+3)(7)?

hartnn (hartnn):

O.o

OpenStudy (anonymous):

hmm, I've forgotten what happens when you factor out, does it yield an x or a 1?

hartnn (hartnn):

what is common from (x+2)^7 and (x+2)^8 ?

OpenStudy (anonymous):

the (x+2)?

hartnn (hartnn):

only x+2 ?

hartnn (hartnn):

\(\Large ☺^3 + ☺^5 = ☺^3 + ☺^3 ☺^2 =☺^3(1+☺^2) \) does that makes sense ?

OpenStudy (anonymous):

wait a moment

OpenStudy (anonymous):

(x+2)^15 would combine them

hartnn (hartnn):

why would u combine them ?

OpenStudy (anonymous):

ok, I guess it doesn't make sense, I am just remembering these rules with powers

OpenStudy (anonymous):

yes, it makes total sense

hartnn (hartnn):

\(\Large (x+2)^7 + (x+2)^8 = (x+2)^7(1+(x+2))\) sure ?

hartnn (hartnn):

in same way, what is common from (x+3)^6 and (x+3)^5 ?

OpenStudy (anonymous):

((x+3)^5)(1+(x+3))

hartnn (hartnn):

correct so, 8(x+2)^7 (x+3)^6 + (x+2)^8 6(x+3)^5 =...?

OpenStudy (anonymous):

hmm

OpenStudy (anonymous):

ok writing down

OpenStudy (anonymous):

so, it's (2)(x+2)^7(x+3)^5 ((4x+12)+(3x+6))

hartnn (hartnn):

good!

OpenStudy (anonymous):

=(2)(x+2)^7(x+3)^5 (7x+18)

hartnn (hartnn):

you are \(\huge \color{red}{\checkmark}\)

OpenStudy (anonymous):

^_^ is that good enough of an answer. I don't have to simplify this one they say

hartnn (hartnn):

thats the most simplified form :)

OpenStudy (anonymous):

woohoo! :) I appreciate the help

hartnn (hartnn):

welcome ^_^

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