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help? step by step Find dy/dx

Mathematics
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\[y=(12\sqrt[3]{x}-\frac{ 1 }{ 4x }+e^2)\]
I know I should rewrite it out.. 12*x^(1/3) but now I'm stuck..?
\[\sqrt[3]{x} = x^(\frac{ 1 }{ 3 }\] by the property, you subtract 1. So 1/3 - 1 = ? \[\frac{ 1 }{ 4x } = \frac{ 1 }{ 4 }x^{-1} \] and \[e^2 \] is just a constant.

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Other answers:

\[x^{\frac{ 1 }{ 3 }}\]
okay, thanks.
\[y' = \frac{ 4 }{ x^{\frac{ 2 }{ 3 } }} - \frac{ 1 }{ 4x^2 }\]
so.. rewritten it's \[(12*x^{1/3}-\frac{ 1 }{ 4 }x^{-1}+e^2)\]
sorry, change the - to +.
ahhh i messed up on the original equation
\[(12\sqrt[3]{x}-\frac{ 1 }{ 4x }+e^2)^7\]
so it's chain rule right?
It's the same thing. now you apply chain rule.
\[7(4x^{\frac{ -2 }{ 3 }} + \frac{ 1 }{ 4x^2 })^6\]
mm.. the answer is different then that.
\[7(12\sqrt[3]{x}-\frac{ 1 }{ 4x }+e^2)^6(4x^{-2/3}+\frac{ 1 }{4 }x^{-2})\]
oh yeah, you multiply by the deriv again.
why do you multiply by the derivative?
oh nevermind haha
I got it.
I'm forgetting my simple rulessss! Ahhhh D;
chain rule states f'(g(x))*g'(x)
it's all good, i totally 4got chain rule for a moment. im rusty in my math
thanks so much though ;D
i remember the chain rule by ths let u equal some function within a function and y=\[f(u)\] so you have\[y=f(u)\] \[\frac{dy}{dx}=\frac{du}{dx}\frac{dy}{du}\] if you remember one party like du/dx or dy/du.... you'll know what you ned to cancel
|dw:1355374308512:dw|
\[(x-2)^x\]
^ I figured it out lol
chain rule?
i've never seen this lol.... suprisingly
i would've just chain ruled it
it's x-2^{x}
=1-2^{x}ln(2)
wat? Lol
nothing abb0t lol :)
is that an integration rule?
o gahd not the ln way of finding derivatives -.-
lmao, ya --> ;/
wait for ODE for differentiation w/ integration.
what do you mean? i've taken Diff Eq
suprisingly i've gone all the way through math not seeing anything like that lol
or i did and it was a longg time ago
for exact equations. also in PDE. chain rule.
exact dquations are simple, \[Ndy=Mdx\] or i might have the backwards i always seem to do it

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