jcamargo93
Find dy/dx. y=5^(ln(x^2)+3)



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abb0t
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\[a^x \ln(a)\]

jcamargo93
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thank you!:)

abb0t
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fa sho! GL!

RadEn
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lol, the clue incomplete...
use chain rule :
y=x^2 > y'=2x
v=ln(u) > v'=1/u
w=5^v > w'=5^v * ln(5)

jcamargo93
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ok, i got this.. y'= 5^(ln((x^2)+3)) *ln 5*((2x)/((x^2)+3)) ..is it right?

RadEn
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for the power of 5, is
{ln(x^2) + 3} or ln(x^2+3)

jcamargo93
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i did include that, im just not sure if i did it right?, do i have to put (1/5) for ln5?

RadEn
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before verify, i want to asking u is ur question
dw:1355302758870:dw

RadEn
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which one, the 1st or the 2nd ?

jcamargo93
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the 2nd one

RadEn
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hohohoo... ok,, u have typo above, right ?

jcamargo93
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i think so,..

jcamargo93
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but yea, my question is to derive the 2nd one

RadEn
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if u have typed ur question like the original above, the result will be difference :)
OK,,,, now U RE RIGHT :*

jcamargo93
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so im right?

RadEn
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sure... nice answer.

RadEn
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but if ur question y=5^(ln(x^2)+3), so the answer will be
y' = (2/x) * 5^(ln(x^2)+3) * ln(5)

jcamargo93
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ok then, well its like the second one

RadEn
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yea...
be careful in type the equation ^^

jcamargo93
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ok thanks!:)

RadEn
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very welcome