anonymous
  • anonymous
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)]
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
jamiebookeater
  • jamiebookeater
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
this 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
  • idku
you have to know that: \(\ln(e^x)=x\)
anonymous
  • anonymous
I am trying to find the derivative of all of this as a whole ln[(e^z)(cos(z))+4π3)], so I'm confused

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

idku
  • idku
ln((e^z)(cos(z))+4π3)) ln(e^z) +\ln(cos(z))+4π3)) z + \ln(cos(z))+4π3)) z - sin(z)/(cos(z))+4π3)
idku
  • idku
ln(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
  • idku
i subtract sin, because: d/dx cos(x) = -sin(x)
idku
  • idku
oh I should have 1 - sin(z)/(cos(z))+4π3)
idku
  • idku
deriv of z is 1, I didn't write that sorry
anonymous
  • anonymous
I'm not really following that train of thought, but thank you! Can anyone else try to explain?
anonymous
  • anonymous
\[\ln [e^z \left( \cos z+4 \pi^3 \right)]=\ln e^z+\ln \left( \cos z+4 \pi ^3 \right)=z+\ln \left( \cos z+ 4\pi^3 \right)\]
anonymous
  • anonymous
its derivative \[=z \prime+\frac{ -\sin z~z \prime }{\cos z+4 \pi ^3 }\]
anonymous
  • anonymous
@Hero
zepdrix
  • zepdrix
So you would like to take the derivative of the entire thing `without first apply log rules`?
anonymous
  • anonymous
It just told me to find the derivative of ln[(e^z)(cos(z))+4π3)] in which the fashion of ln(x)
zepdrix
  • zepdrix
So 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
  • zepdrix
Ah I messed up my brackets :p woops
zepdrix
  • zepdrix
\[\large\rm \frac{d}{dx}\ln\left[\color{#CC0033}{(e^z(\cos z)+4π^3)}\right]=\frac{1}{\color{#CC0033}{(e^z(\cos z)+4π^3)}}\cdot\frac{d}{dx}\color{#CC0033}{(e^z(\cos z)+4π^3)}\]
zepdrix
  • zepdrix
Don'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
  • zepdrix
No power rule for him. Derivative of a constant is zero.
zepdrix
  • zepdrix
Here 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
  • zepdrix
Oh obviously that operator should be d/dz, my mistake :(
zepdrix
  • zepdrix
Taking 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
  • zepdrix
Ah ok great! Your answer looks correct besides the 12pi^2 portion :)
zepdrix
  • zepdrix
Do you understand why that disappeared?
anonymous
  • anonymous
Oh my goodness yes! This makes so much sense now! Thank you!
zepdrix
  • zepdrix
yay team \c:/

Looking for something else?

Not the answer you are looking for? Search for more explanations.