log differential

[Scott92]

I have the following question but two ways to solve it not sure which if either is correct
Differentiate each of the following voltage functions with respect to time and hence determine the ‘rate of change’ for each of the functions when time (t) is 5 seconds v=loge(2t)
loge(2×5)

loge(2×5)=loge(2)+loge+(5) Use log rule loge(x)=In(x)

In(2)+loge(5)=In(2)+In(5)=2.302585093 vs

0r simply 1/t 1/5=0.2vs

 

[Dr.Peterson]

I have the following question but two ways to solve it not sure which if either is correct
Differentiate each of the following voltage functions with respect to time and hence determine the ‘rate of change’ for each of the functions when time (t) is 5 seconds v=log_e(2t)
log_e(2×5)

log_e(2×5)=log_e(2)+log_e+(5) Use log rule log_e(x)=In(x)

In(2)+log_e(5)=In(2)+In(5)=2.302585093 vs

0r simply 1/t 1/5=0.2vs

Only the last line looks sort of correct, but you haven't supported it.

What is the derivative of [imath]f(t)=\ln(2t)[/imath]? Don't replace t with 5 until after you've found [imath]f'(t)[/imath].

Have you tried using the chain rule? You can also simplify first, as I think you are trying to do, and not need the chain rule; but you have to do that correctly.

 

[Scott92]

Hi thanks for helping
In the work book i have all it gives me about logs is function A loge (x) Derivative being A/x with a given example 16loge (x) as 16/x but doesn't really explain in detail so little confused with it

 

[Dr.Peterson]

Hi thanks for helping
In the work book i have all it gives me about logs is function A loge (x) Derivative being A/x with a given example 16loge (x) as 16/x but doesn't really explain in detail so little confused with it

So you haven't learned the chain rule? What have you been taught about derivatives? (Incidentally, this topic would be called calculus; "differential equations" is a more advanced topic using derivatives.)

But that doesn't matter. You can simplify [imath]\ln(2t)[/imath] using the product rule for logs, and then differentiate it. Please show your work for that.

 

[Steven G]

ln(2t) = ln(2) + ln(t).
Now use what your workbook told you and differentiate the above. Then and only then do you put in the value for t.

 

[Scott92]

Thanks will look up chain rule as the work book just gives examples its not the best,

So it would be In(2)+In(5) In2/In5=0.4306765581vs ??

 

[Deleted member 4993]

Thanks will look up chain rule as the work book just gives examples its not the best,

So it would be In(2)+In(5) In2/In5=0.4306765581vs ??

No! Look at response #5 carefully!

V(t) = ln(2t) = ln(2) + ln(t)

V'(t) = 0 + (1/t)

V'(5) = ??

 

[Steven G]

Thanks will look up chain rule as the work book just gives examples its not the best,

So it would be In(2)+In(5) In2/In5=0.4306765581vs ??

ln5 - ln2 = ln(5/2)

 

[Scott92]

No! Look at response #5 carefully!

V(t) = ln(2t) = ln(2) + ln(t)

V'(t) = 0 + (1/t)

V'(5) = ??

Id guess at 1/5 but i don't really understand it trying get back into high level maths after 10 plus years is pretty rough will try watch some videos on product rules and log rules and try make it stick Thanks

 

[Deleted member 4993]

i don't really understand it

In which step did you get lost?

Do you know (remember):

\(\displaystyle \frac{d}{dx}[log_e(x)] \ = \ \frac{1}{x} \)

 

[Scott92]

In which step did you get lost?

Do you know (remember):

\(\displaystyle \frac{d}{dx}[log_e(x)] \ = \ \frac{1}{x} \)

Yeah I get that but with it being log 2t I don't get what I do with the 2 as most examples are just a letter without a number by it or a squared letter

 

[Deleted member 4993]

Yeah I get that but with it being log 2t I don't get what I do with the 2 as most examples are just a letter without a number by it or a squared letter

Please post an example of those "most examples".

 

[Steven G]

You have two options.

1) Think of ln(2t) = ln(2) + ln(t). Now \(\displaystyle \dfrac{d}{dt}(ln(2t)) = \dfrac{d}{dt}(ln(2) + ln(t))=\dfrac{d}{dt}(ln(2)) + \dfrac{d}{dt}(ln(t))\)
You continue from here.

2) Using the chain rule. \(\displaystyle \dfrac{d}{dt}(ln(u)) = \dfrac{u'}{u}\). In your example, u=2t. Finish up

 

[Scott92]

You have two options.

1) Think of ln(2t) = ln(2) + ln(t). Now \(\displaystyle \dfrac{d}{dt}(ln(2t)) = \dfrac{d}{dt}(ln(2) + ln(t))=\dfrac{d}{dt}(ln(2)) + \dfrac{d}{dt}(ln(t))\)
You continue from here.

2) Using the chain rule. \(\displaystyle \dfrac{d}{dt}(ln(u)) = \dfrac{u'}{u}\). In your example, u=2t. Finish up

Number one 1/t ?
Number two 1/t /2t
So if that's correct do I then multiply 2 and the 5 over the 1/2*5 ? To get a figure

 

[Steven G]

You didn't follow the formula for number two. You need to divide u' by u. So 1st you need to find u and then u'.

Are you suggesting that there are two different answers.
You need to know what equal signs really mean!
If ln(2t) = ln(2) + ln(t), the the derivative of the left hand side (my number 2 above) must equal the derivative of the right hand side (my number 1 above)