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monroe17Best ResponseYou've already chosen the best response.1
\[y=(12\sqrt[3]{x}\frac{ 1 }{ 4x }+e^2)\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
I know I should rewrite it out.. 12*x^(1/3) but now I'm stuck..?
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
\[\sqrt[3]{x} = x^(\frac{ 1 }{ 3 }\] 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.
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
\[y' = \frac{ 4 }{ x^{\frac{ 2 }{ 3 } }}  \frac{ 1 }{ 4x^2 }\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
so.. rewritten it's \[(12*x^{1/3}\frac{ 1 }{ 4 }x^{1}+e^2)\]
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
sorry, change the  to +.
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
ahhh i messed up on the original equation
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
\[(12\sqrt[3]{x}\frac{ 1 }{ 4x }+e^2)^7\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
so it's chain rule right?
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
It's the same thing. now you apply chain rule.
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
\[7(4x^{\frac{ 2 }{ 3 }} + \frac{ 1 }{ 4x^2 })^6\]
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
mm.. the answer is different then that.
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
\[7(12\sqrt[3]{x}\frac{ 1 }{ 4x }+e^2)^6(4x^{2/3}+\frac{ 1 }{4 }x^{2})\]
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
oh yeah, you multiply by the deriv again.
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
why do you multiply by the derivative?
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
I'm forgetting my simple rulessss! Ahhhh D;
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
chain rule states f'(g(x))*g'(x)
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
it's all good, i totally 4got chain rule for a moment. im rusty in my math
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
thanks so much though ;D
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
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
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
dw:1355374308512:dw
 one year ago

monroe17Best ResponseYou've already chosen the best response.1
^ I figured it out lol
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
i've never seen this lol.... suprisingly
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
i would've just chain ruled it
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
is that an integration rule?
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
o gahd not the ln way of finding derivatives .
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
wait for ODE for differentiation w/ integration.
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
what do you mean? i've taken Diff Eq
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
suprisingly i've gone all the way through math not seeing anything like that lol
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
or i did and it was a longg time ago
 one year ago

abb0tBest ResponseYou've already chosen the best response.1
for exact equations. also in PDE. chain rule.
 one year ago

Outkast3r09Best ResponseYou've already chosen the best response.0
exact dquations are simple, \[Ndy=Mdx\] or i might have the backwards i always seem to do it
 one year ago
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