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monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1\[y=(12\sqrt[3]{x}\frac{ 1 }{ 4x }+e^2)\]

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1I know I should rewrite it out.. 12*x^(1/3) but now I'm stuck..?

abb0t
 2 years ago
Best 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.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1\[y' = \frac{ 4 }{ x^{\frac{ 2 }{ 3 } }}  \frac{ 1 }{ 4x^2 }\]

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1so.. rewritten it's \[(12*x^{1/3}\frac{ 1 }{ 4 }x^{1}+e^2)\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1sorry, change the  to +.

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1ahhh i messed up on the original equation

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1\[(12\sqrt[3]{x}\frac{ 1 }{ 4x }+e^2)^7\]

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1so it's chain rule right?

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1It's the same thing. now you apply chain rule.

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1\[7(4x^{\frac{ 2 }{ 3 }} + \frac{ 1 }{ 4x^2 })^6\]

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1mm.. the answer is different then that.

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1\[7(12\sqrt[3]{x}\frac{ 1 }{ 4x }+e^2)^6(4x^{2/3}+\frac{ 1 }{4 }x^{2})\]

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1oh yeah, you multiply by the deriv again.

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1why do you multiply by the derivative?

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1I'm forgetting my simple rulessss! Ahhhh D;

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1chain rule states f'(g(x))*g'(x)

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1it's all good, i totally 4got chain rule for a moment. im rusty in my math

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1thanks so much though ;D

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0i 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

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0dw:1355374308512:dw

monroe17
 2 years ago
Best ResponseYou've already chosen the best response.1^ I figured it out lol

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0i've never seen this lol.... suprisingly

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0i would've just chain ruled it

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0is that an integration rule?

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0o gahd not the ln way of finding derivatives .

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1wait for ODE for differentiation w/ integration.

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0what do you mean? i've taken Diff Eq

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0suprisingly i've gone all the way through math not seeing anything like that lol

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0or i did and it was a longg time ago

abb0t
 2 years ago
Best ResponseYou've already chosen the best response.1for exact equations. also in PDE. chain rule.

Outkast3r09
 2 years ago
Best ResponseYou've already chosen the best response.0exact dquations are simple, \[Ndy=Mdx\] or i might have the backwards i always seem to do it
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