Solving calculus, how dose this look?

[Mathmasteriw]

Hi Guys and gals,
Is this correct?
Is there a better way to write it out?
Thanks ?

 

[Harry_the_cat]

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]

O great ! Yes that looks a much better way to present the answer!
Thank you Harry! ?

 

[Deleted member 4993]

Hi Guys and gals,
Is this correct?
Is there a better way to write it out?
Thanks ?
View attachment 24279

V(t) is NOT a function of 'x' !

Why do you have 'dV/dx' in your work ?!!

 

[lookagain]

Also, consider using $\displaystyle \ V \ = \ log_e(2t) \ = \ log_e(2) \ + \ log_e(t) \$ if you don't have to use the chain rule.

 

[Steven G]

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.

ALog_eu-->(A/u)du/dx

 

[Steven G]

This is another formula you should know. $$\dfrac{d}{dx}(A\log_e{kx}) = A/x$$. This will make your life easier for the problem you are working on.

 

[Mathmasteriw]

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.

ALog_eu-->(A/u)du/dx

Thanks Jomo!!