Solving calculus, how dose this look?

Mathmasteriw

Junior Member
Joined
Oct 22, 2020
Messages
83
Hi Guys and gals,
Is this correct?
Is there a better way to write it out?
Thanks 😘
2FFACDB2-76EC-41AC-A931-FB6125E8679C.jpeg
 

Harry_the_cat

Senior Member
Joined
Mar 16, 2016
Messages
2,359
You could write it out using the chain rule.

\(\displaystyle V= log_e(2t)\)

Let \(\displaystyle V=log_e(u)\) where \(\displaystyle u=2t\)

Therefore \(\displaystyle \frac{dV}{du}=\frac{1}{u}\) and \(\displaystyle \frac{du}{dt}=2\)

\(\displaystyle \frac{dV}{dt}=\frac{dV}{du}\frac{du}{dt} = \frac{1}{u} *2=\frac{2}{u} = \frac{2}{2t} = \frac{1}{t}\)
 

Mathmasteriw

Junior Member
Joined
Oct 22, 2020
Messages
83
O great ! Yes that looks a much better way to present the answer!
Thank you Harry! 😺
 

lookagain

Elite Member
Joined
Aug 22, 2010
Messages
2,680
Also, consider using \(\displaystyle \ V \ = \ log_e(2t) \ = \ log_e(2) \ + \ log_e(t) \ \) if you don't have to use the chain rule.
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
9,669
I will only suggest this one more time. I think that the formulas you are using from your book is causing you trouble. I advised you to update all of them. Simply replace x with u on the function side and on the derivative side replace x with u and multiply by du/dx.

ALogeu-->(A/u)du/dx
 

Jomo

Elite Member
Joined
Dec 30, 2014
Messages
9,669
This is another formula you should know. \(\displaystyle \dfrac{d}{dx}(A\log_e{kx}) = A/x\). This will make your life easier for the problem you are working on.
 

Mathmasteriw

Junior Member
Joined
Oct 22, 2020
Messages
83
I will only suggest this one more time. I think that the formulas you are using from your book is causing you trouble. I advised you to update all of them. Simply replace x with u on the function side and on the derivative side replace x with u and multiply by du/dx.

ALogeu-->(A/u)du/dx
Thanks Jomo!!
 
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