# Solving calculus, how dose this look?

#### Mathmasteriw

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

#### Harry_the_cat

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

#### Subhotosh Khan

##### Super Moderator
Staff member
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

##### Elite Member
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
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
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
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!!