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anonymous
 one year ago
Find the derivative:
ln((e^z)(cos(z))+4π3))
I got this for my answer, but apparently it is incorrect.
[(e^z)*([sin(z)])+cos(z)*(e^z)+12*(pi^2)]/[(e^z)*cos(z)+4*(pi^3)]
anonymous
 one year ago
Find the derivative: ln((e^z)(cos(z))+4π3)) I got this for my answer, but apparently it is incorrect. [(e^z)*([sin(z)])+cos(z)*(e^z)+12*(pi^2)]/[(e^z)*cos(z)+4*(pi^3)]

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0this is my answer in more of a equation form \[[(e^z)*([\sin(z)])+\cos(z)*(e^z)+12*(\pi^2)]/[(e^z)*\cos(z)+4*(\pi^3)]\]

idku
 one year ago
Best ResponseYou've already chosen the best response.2you have to know that: \(\ln(e^x)=x\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I am trying to find the derivative of all of this as a whole ln[(e^z)(cos(z))+4π3)], so I'm confused

idku
 one year ago
Best ResponseYou've already chosen the best response.2ln((e^z)(cos(z))+4π3)) ln(e^z) +\ln(cos(z))+4π3)) z + \ln(cos(z))+4π3)) z  sin(z)/(cos(z))+4π3)

idku
 one year ago
Best ResponseYou've already chosen the best response.2ln(ab)=ln(a)+ln(b) then ln(e^z)=z, and that is why I have z+ ... and then chain rule for ln(cos(z))+4π3)

idku
 one year ago
Best ResponseYou've already chosen the best response.2i subtract sin, because: d/dx cos(x) = sin(x)

idku
 one year ago
Best ResponseYou've already chosen the best response.2oh I should have 1  sin(z)/(cos(z))+4π3)

idku
 one year ago
Best ResponseYou've already chosen the best response.2deriv of z is 1, I didn't write that sorry

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'm not really following that train of thought, but thank you! Can anyone else try to explain?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln [e^z \left( \cos z+4 \pi^3 \right)]=\ln e^z+\ln \left( \cos z+4 \pi ^3 \right)=z+\ln \left( \cos z+ 4\pi^3 \right)\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0its derivative \[=z \prime+\frac{ \sin z~z \prime }{\cos z+4 \pi ^3 }\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So you would like to take the derivative of the entire thing `without first apply log rules`?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0It just told me to find the derivative of ln[(e^z)(cos(z))+4π3)] in which the fashion of ln(x)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1So you're trying to apply this rule I guess: \[\large\rm \frac{d}{dx}\ln(\color{#CC0033}{x})=\frac{1}{\color{#CC0033}{x}}\]But we'll also need to chain rule :) \[\large\rm \frac{d}{dx}\ln\left[\color{#CC0033}{e^z(\cos z)+4π^3)}\right]=\frac{1}{\color{#CC0033}{e^z(\cos z)+4π^3)}}\cdot\frac{d}{dx}\color{#CC0033}{e^z(\cos z)+4π^3)}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ah I messed up my brackets :p woops

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1\[\large\rm \frac{d}{dx}\ln\left[\color{#CC0033}{(e^z(\cos z)+4π^3)}\right]=\frac{1}{\color{#CC0033}{(e^z(\cos z)+4π^3)}}\cdot\frac{d}{dx}\color{#CC0033}{(e^z(\cos z)+4π^3)}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Don't let the 4pi^3 fool you! It's a fancy fancy constant, dressed up in a tuxedo and all that jazz. He's trying to play a trick on you.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1No power rule for him. Derivative of a constant is zero.

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Here is our setup for product rule:\[\large\rm \frac{d}{dx}(e^z(\cos z)+4π^3)=\color{royalblue}{(e^z)'}(\cos z)+e^z\color{royalblue}{(\cos z)'}+\frac{d}{dx}4\pi^3\]As mentioned before, derivative of a constant is zero,\[\large\rm \frac{d}{dx}(e^z(\cos z)+4π^3)=\color{royalblue}{(e^z)'}(\cos z)+e^z\color{royalblue}{(\cos z)'}+0\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Oh obviously that operator should be d/dz, my mistake :(

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Taking derivatives:\[\large\rm \frac{d}{dz}(e^z(\cos z)+4π^3)=\color{orangered}{(e^z)}(\cos z)+e^z\color{orangered}{(\sin z)}\]

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Ah ok great! Your answer looks correct besides the 12pi^2 portion :)

zepdrix
 one year ago
Best ResponseYou've already chosen the best response.1Do you understand why that disappeared?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Oh my goodness yes! This makes so much sense now! Thank you!
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