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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)]

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  1. anonymous
    • one year ago
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    this is my answer in more of a equation form \[[(e^z)*(-[\sin(z)])+\cos(z)*(e^z)+12*(\pi^2)]/[(e^z)*\cos(z)+4*(\pi^3)]\]

  2. idku
    • one year ago
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    you have to know that: \(\ln(e^x)=x\)

  3. anonymous
    • one year ago
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    I am trying to find the derivative of all of this as a whole ln[(e^z)(cos(z))+4π3)], so I'm confused

  4. idku
    • one year ago
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    ln((e^z)(cos(z))+4π3)) ln(e^z) +\ln(cos(z))+4π3)) z + \ln(cos(z))+4π3)) z - sin(z)/(cos(z))+4π3)

  5. idku
    • one year ago
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    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)

  6. idku
    • one year ago
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    i subtract sin, because: d/dx cos(x) = -sin(x)

  7. idku
    • one year ago
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    oh I should have 1 - sin(z)/(cos(z))+4π3)

  8. idku
    • one year ago
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    deriv of z is 1, I didn't write that sorry

  9. anonymous
    • one year ago
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    I'm not really following that train of thought, but thank you! Can anyone else try to explain?

  10. anonymous
    • one year ago
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    \[\ln [e^z \left( \cos z+4 \pi^3 \right)]=\ln e^z+\ln \left( \cos z+4 \pi ^3 \right)=z+\ln \left( \cos z+ 4\pi^3 \right)\]

  11. anonymous
    • one year ago
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    its derivative \[=z \prime+\frac{ -\sin z~z \prime }{\cos z+4 \pi ^3 }\]

  12. anonymous
    • one year ago
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    @Hero

  13. zepdrix
    • one year ago
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    So you would like to take the derivative of the entire thing `without first apply log rules`?

  14. anonymous
    • one year ago
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    It just told me to find the derivative of ln[(e^z)(cos(z))+4π3)] in which the fashion of ln(x)

  15. zepdrix
    • one year ago
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    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 :) \[\large\rm \frac{d}{dx}\ln\left[\color{#CC0033}{e^z(\cos z)+4π^3)}\right]=\frac{1}{\color{#CC0033}{e^z(\cos z)+4π^3)}}\cdot\frac{d}{dx}\color{#CC0033}{e^z(\cos z)+4π^3)}\]

  16. zepdrix
    • one year ago
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    Ah I messed up my brackets :p woops

  17. zepdrix
    • one year ago
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    \[\large\rm \frac{d}{dx}\ln\left[\color{#CC0033}{(e^z(\cos z)+4π^3)}\right]=\frac{1}{\color{#CC0033}{(e^z(\cos z)+4π^3)}}\cdot\frac{d}{dx}\color{#CC0033}{(e^z(\cos z)+4π^3)}\]

  18. zepdrix
    • one year ago
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    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.

  19. zepdrix
    • one year ago
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    No power rule for him. Derivative of a constant is zero.

  20. zepdrix
    • one year ago
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    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,\[\large\rm \frac{d}{dx}(e^z(\cos z)+4π^3)=\color{royalblue}{(e^z)'}(\cos z)+e^z\color{royalblue}{(\cos z)'}+0\]

  21. zepdrix
    • one year ago
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    Oh obviously that operator should be d/dz, my mistake :(

  22. zepdrix
    • one year ago
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    Taking derivatives:\[\large\rm \frac{d}{dz}(e^z(\cos z)+4π^3)=\color{orangered}{(e^z)}(\cos z)+e^z\color{orangered}{(-\sin z)}\]

  23. zepdrix
    • one year ago
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    Ah ok great! Your answer looks correct besides the 12pi^2 portion :)

  24. zepdrix
    • one year ago
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    Do you understand why that disappeared?

  25. anonymous
    • one year ago
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    Oh my goodness yes! This makes so much sense now! Thank you!

  26. zepdrix
    • one year ago
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    yay team \c:/

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