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Maths wizards, remind me is there a derivative of 2^3x.dx ?

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Y=2^(3x). Ln both sides.
Lny=ln 2^(3x). Use prop of logs.
Lny=3x (ln2). Diff each side.
1/y dy/dx= 3(ln2). Solve for dy/dx.
dy/dx= 3(ln2)(y). Sub orig equation in.
dy/dx= 3(ln2)(2^3x)

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Yes that is very neat...
Would not have thought of making ln of both sides to "set it up"
Very well explained
You're an expert hehe.
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Can you help the fellow asking the integral of 2^3x.3^2x.5^x
I imagine it would be integration by parts
and integration by parts again on one of those parts
5am here heh
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Oh, I wish I could do it! I'll try to get the answer!
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Uhhhhh... Guess I'm not a math wizard.
I'll stick with my history...

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I'm thinking

if u=3x and use the chain rule...

and I have y = 2^u

dy/dx = dy/du x du/dx

du/dx = 3

but

dy/du would be ? ?

hmm perhaps history is safer huh ? hehe.
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Yup.
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y=2^3X
Y ' = 2^3X ln2
= 2^3X x 0.6
let me know if you don't get it
I'm a vampire by the way ;)

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Can you give me a link that shows that Y ' = 2^3X ln2 ?

Is it the chain rule ?

I believe the integral is (int)Y' = 2^3X / ln2

Can you solve the integral of (2^3x).(3^2x).(5^x) ?
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Whats the anagram of VAMPIRE and VAMPIRE + AL ?

VAMPIER
PRIMAEVAL and VERAPAMIL a medicinal drug (a vasodilator)
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no it's not the chain rule. it's derivative of exponential function.
y=a^x , a>0
y ' = a^x lna
i looked for it on the internet and i found this:
http://tutorial.math.lamar.edu/Classes/CalcI/DiffExpLogFcns.aspx
and lol I'd never thought of this anagram :D
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I knew a good website about math when i was in high school but now unfortunately I've forgotten :(
but I'm sure this derivative is correct.
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Wow thats brilliant (and correct)

I searched but couldn't find it anywhere
Thanks so much.
Yee ha
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Can you save me the effort of dy/dx of a^bx.dx ?
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implicit differentiation? you don't need to do that cause there's an explicit function.
like the other one y ' = b x a^bX lna
because:
y=a^u , a>0
y ' = u ' a^u lna
sorry for the late response. i wasn't home.
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Between you and Mary I have understood completely
just had to reawake rusty maths skills,
and you both helped enormously.
thanks so much.
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you're welcome :)
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