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