Help with Function

[Mathmasteriw]

hi there!
Hopeing you can help me with this...
I just want to check if both these functions are the same?
I presume 1loge is the same as loge? Just want to double check!
Thanks for your time

 

[HallsofIvy]

I wondered what a "loge" (a theater seat) was doing in a math problem! This is the "logarithm base e", $\displaystyle log_e(2x)$ or $\displaystyle ln(2x)$. Yes, 1 times anything is just that thing. But that would be very peculiar to write. Are you certain that is what you are reading?

 

[Mathmasteriw]

 

[Mathmasteriw]

A yes apologies for my amature approach to presenting the problem, my bad im new to this advanced stuff.
Heres the question on top of the last photo,
And this is how ive derived it. Any good?

 

[Dr.Peterson]

A yes apologies for my amature approach to presenting the problem, my bad im new to this advanced stuff.
Heres the question on top of the last photo,
And this is how ive derived it. Any good?

I think what you are saying is that you were given the function [MATH]v = \log_e(2t)[/MATH] and want to find its derivative at [MATH]t = 5[/MATH]. (We say you are "differentiating" it, not "deriving" it; that's an oddity of English math terminology I wish we could fix!)

You evidently have been given a formula that if [MATH]y = A\log_e(x)[/MATH], then [MATH]\frac{dy}{dx} = \frac{A}{x}[/MATH]. (You didn't state what function the formula is meant to apply to, but this is what would fit.)

Applying this formula to your function, taking [MATH]x = 2t[/MATH] and [MATH]A = 1[/MATH], you get [MATH]\frac{dv}{dt} = \frac{1}{2t}[/MATH]. But if so, you forgot to apply the chain rule.

I may be misinterpreting what you intend, because your notation is quite wrong, particularly in writing "=" between a function and its derivative. It will help if you show the actual problem you are working on (even if it's in another language), and the formula(s) you are applying, so we can be sure what you mean.

 

[Mathmasteriw]

I think what you are saying is that you were given the function [MATH]v = \log_e(2t)[/MATH] and want to find its derivative at [MATH]t = 5[/MATH]. (We say you are "differentiating" it, not "deriving" it; that's an oddity of English math terminology I wish we could fix!)

You evidently have been given a formula that if [MATH]y = A\log_e(x)[/MATH], then [MATH]\frac{dy}{dx} = \frac{A}{x}[/MATH]. (You didn't state what function the formula is meant to apply to, but this is what would fit.)

Applying this formula to your function, taking [MATH]x = 2t[/MATH] and [MATH]A = 1[/MATH], you get [MATH]\frac{dv}{dt} = \frac{1}{2t}[/MATH]. But if so, you forgot to apply the chain rule.

I may be misinterpreting what you intend, because your notation is quite wrong, particularly in writing "=" between a function and its derivative. It will help if you show the actual problem you are working on (even if it's in another language), and the formula(s) you are applying, so we can be sure what you mean.

Thank you so much for your time and your knowledge is greatly appreciated Dr.
I see thanks for those pointers,
Here’s a clearer version of the full question.
Thank you

 

[Steven G]

Hi,
You clearly wrote that dv/dt = 1/10 which is not true. dv/dt is Not constant. Maybe dv/dt evaluated at t= 5, written as dv/dt(5) or dv/dt |_t=5, is a constant but dv/dt is not. Also you are not using the chain rule. Please try again and remember that equal signs must be valid.

 

[Steven G]

Perhaps your formula is not friendly enough.

Maybe this will be better. $$A\dfrac{d}{dt}(\log_eu)=\dfrac{A}{u}*\dfrac{du}{dx}$$

 

[JeffM]

You are missing the point. We want to see a picture of what was given to you, NOT a picture of what you have written down. It is hard to be sure what you mean. t = 5r? What's r supposed to be? log e is mathematically meaningless. I suspect you mean log_e, but I cannot be sure. Is that (2b) or (26)? What you are sending us is a mess. It is not what we want.

 

[Steven G]

You are missing the point. We want to see a picture of what was given to you, NOT a picture of what you have written down. It is hard to be sure what you mean. t = 5r? What's r supposed to be? log e is mathematically meaningless. I suspect you mean log_e, but I cannot be sure. Is that (2b) or (26)? What you are sending us is a mess. It is not what we want.

Jeff, I agree that we want the OP to post the exact problem. As far as 5? goes I suspect it is 5s as in 5 seconds.

 

[Dr.Peterson]

Thank you so much for your time and your knowledge is greatly appreciated Dr.
I see thanks for those pointers,
Here’s a clearer version of the full question.
Thank you

Amid the discussion of poor notation, and wanting to see an image of the problem itself as given to you in print, please don't miss the important fact:

You need to apply the chain rule when you differentiate, so your answer so far is wrong. Please make an attempt to correct this.

 

[Mathmasteriw]

Hi everyone, thank you all for all feedback!
Sory , been busy last couple of days, so here’s a photo of the original problem and of the table of derivatives.
I will look again on the chain rule.
Thanks

 

[HallsofIvy]

Okay, you want to find the derivative of $\displaystyle log_e(2t)$. First I hope you understand that whether the variable is called "t" or "x" or anything else is not important. We just use whatever letter the problem uses.

Your table tells you that the derivative of "$\displaystyle Aln_e(x)$" is $\displaystyle \frac{A}{x}$. I would add a general rule to your table: The derivative of f+ g, where f and g are two functions is the derivative of f plus the derivative of g: $\displaystyle \frac{d(f+ g)}{dx}= \frac{df}{dx}+ \frac{dg}{dx}$ (or, equivalently, $\displaystyle \frac{d(f+ g)}{dt}= \frac{df}{dt}+ \frac{dg}{dt}$).

There is NO number multiplying $\displaystyle ln_e( )$ but that just means that A= 1. The derivative of $\displaystyle log_e(x)$ is $\displaystyle \frac{1}{x}$. But what about that "2"?

Well, of course, one of the "properties of logarithms" is that $\displaystyle log(ab)= log(a)+ log(b)$ so that $\displaystyle log_e(2x)= log_e(2)+ log_e(t)$. \(\displaystyle log_e(2)\(\displaystyle is a constant! And though it is not included in your table of derivatives, the derivative of any constant function is 0. The derivative of $\displaystyle log_e(2t)$ is just the derivative of $\displaystyle log_e(t)$ which your table tells you is $\displaystyle \frac{1}{t}$.

(The "chain rule" would tell you that the derivative of $\displaystyle log_e(2x)$ is $\displaystyle \frac{1}{2x}$ times the derivative of 2x which is 2: $\displaystyle \frac{1}{2x}(2)= \frac{1}{x}$ as before. Given any positive number, a, $\displaystyle log_e(ax)= log_e(a)+ log_e(x)$ so the derivative of $\displaystyle log_e(ax)$ is $\displaystyle \frac{1}{x}$. By the chain rule, the derivative is $\displaystyle \frac{a}{ax}= \frac{1}{x}$.)\)\)