## 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)]

1. anonymous

this is my answer in more of a equation form $[(e^z)*(-[\sin(z)])+\cos(z)*(e^z)+12*(\pi^2)]/[(e^z)*\cos(z)+4*(\pi^3)]$

2. idku

you have to know that: $$\ln(e^x)=x$$

3. 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

4. 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)

5. 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)

6. idku

i subtract sin, because: d/dx cos(x) = -sin(x)

7. idku

oh I should have 1 - sin(z)/(cos(z))+4π3)

8. idku

deriv of z is 1, I didn't write that sorry

9. anonymous

I'm not really following that train of thought, but thank you! Can anyone else try to explain?

10. anonymous

$\ln [e^z \left( \cos z+4 \pi^3 \right)]=\ln e^z+\ln \left( \cos z+4 \pi ^3 \right)=z+\ln \left( \cos z+ 4\pi^3 \right)$

11. anonymous

its derivative $=z \prime+\frac{ -\sin z~z \prime }{\cos z+4 \pi ^3 }$

12. anonymous

@Hero

13. zepdrix

So you would like to take the derivative of the entire thing without first apply log rules?

14. anonymous

It just told me to find the derivative of ln[(e^z)(cos(z))+4π3)] in which the fashion of ln(x)

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

16. zepdrix

Ah I messed up my brackets :p woops

17. zepdrix

$\large\rm \frac{d}{dx}\ln\left[\color{#CC0033}{(e^z(\cos z)+4π^3)}\right]=\frac{1}{\color{#CC0033}{(e^z(\cos z)+4π^3)}}\cdot\frac{d}{dx}\color{#CC0033}{(e^z(\cos z)+4π^3)}$

18. 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.

19. zepdrix

No power rule for him. Derivative of a constant is zero.

20. 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,$\large\rm \frac{d}{dx}(e^z(\cos z)+4π^3)=\color{royalblue}{(e^z)'}(\cos z)+e^z\color{royalblue}{(\cos z)'}+0$

21. zepdrix

Oh obviously that operator should be d/dz, my mistake :(

22. zepdrix

Taking derivatives:$\large\rm \frac{d}{dz}(e^z(\cos z)+4π^3)=\color{orangered}{(e^z)}(\cos z)+e^z\color{orangered}{(-\sin z)}$

23. zepdrix

Ah ok great! Your answer looks correct besides the 12pi^2 portion :)

24. zepdrix

Do you understand why that disappeared?

25. anonymous

Oh my goodness yes! This makes so much sense now! Thank you!

26. zepdrix

yay team \c:/