# Homework Help: Derivative of s(t) = (976(.835)^t -1) + 176t using log differentiation

1. Mar 3, 2012

### LearninDaMath

1. The problem statement, all variables and given/known data

The derivative of s(t) = (976(.835)^t - 1) +176t

I have to take ln of both sides to bring the t down from the exponent. But I never had to apply ln to an equation of this complexity. Here is my attempt, but it doesn't even look close to being on the correct path...what I'm I doing wrong? I'm guessing i'm messing up a step pretty early on in this problem. Can you locate my confusion?

EDIT: the t on the 176t got cut off when scanned, but its supposed to be there on the 3rd and 4th lines..

Last edited: Mar 3, 2012
2. Mar 3, 2012

### Office_Shredder

Staff Emeritus
ln does not do what you want it to do. Once you take ln you're basically done as far as simplifying the equation is concerned. Of course, at that point you could just use the chain rule, but that requires differentiating the original equation as well, which begs the question: why are you taking ln before differentiating if all you want is the derivative of s(t)?

3. Mar 3, 2012

### SammyS

Staff Emeritus
Isolate (on one side of the equation) the 976(.835)t, or (.835)^t before taking the logarithm.

4. Mar 3, 2012

### LearninDaMath

This is the general idea of whre i'm trying to go with this question:

....And i'm trying to fill in the steps that might have been skipped here. All i know about using logs is that when there are variables in the exponent you can take ln of both sides to simplify before differentiating. But i'm not sure if i'm understanding how or when, within this particular problem, to take the ln of both sides.

5. Mar 3, 2012

### LearninDaMath

not exactly sure what u mean. Do u mean that when taking ln of both sides of an equation, all terms should be distributed (so that there are no parenthesis)?

For instance, if I have S = A(B$^{t}$ + C)

And I want to take ln of both sides, I must first do this:

S = AB$^{t}$ +AC

and only then I can take ln:

lnS = ln(AB$^{t}$ + lnAC)

which becomes

LnS = lnAB$^{t}$ + lnAC

would be what you're saying?

6. Mar 3, 2012

### Office_Shredder

Staff Emeritus
The key here isn't to use natural logs, it's to use properties of derivatives. Let me do a simpler example and you can try expanding it to this problem. Calculate the derivative of
$$s(t) = 3*(.5^t)+t$$

$$\frac{d}{dt} \left( 3*(.5^t)+t \right) = \frac{d}{dt} \left(3*.5^t \right) + \frac{d}{dt} \left( t \right)$$
$$= 3 \frac{d}{dt}(.5^t) + \frac{dt}{dt}$$
the derivative of t as a function of t is just t. To differentiate .5t we use the exponential rule $d/dt(a^t) = \ln(a) a^t$ to get
$$= 3*\ln(.5) .5^t+1$$

7. Mar 3, 2012

### SammyS

Staff Emeritus
Another way to say this is:
Solve your equation for 976(.835)t, or (.835)t, before taking the log of both sides.​

Last edited: Mar 3, 2012
8. Mar 3, 2012

### LearninDaMath

So if I have s = .5$^{t}$

if I go right into differentiating, i'll get

ds/dt = t(.5)$^{t-1}$(0) so that

ds/dt = 0

However, if I use the exponenent rule, i'll get

ds/dt = d/dt ln(.5$^{t}$) = ln(.5)*(.5$^{t}$)

Would that be correct?

9. Mar 3, 2012

### SammyS

Staff Emeritus
The point of doing logarithmic differentiation is;
If s = (0.5)t ,

then ln(s) = t ln(0.5) .​

Differentiating that should be easy.

10. Mar 3, 2012

### Office_Shredder

Staff Emeritus
You cannot use the power rule on something of the form at

11. Mar 3, 2012

### LearninDaMath

Okay, then i'm confused at the moment. If you can't use the power rule on a$^{t}$, then how does a$^{t}$ differ from (.5$^{t}$)?

I'm asking because it seems to differ from the state rule:

EDIT: Wait, Wait...by power rule, are referring to chain rule??? If so, then I do understand that power/chain rule can't be applied to a^t because there is a variable as the exponent, right? However, if you are saying that the exponential rule can't be applied to a^t, then I am definately confused.

Last edited: Mar 3, 2012
12. Mar 3, 2012

### LearninDaMath

1. The problem statement, all variables and given/known data

------------------------------------------
So if I have s = .5$^{t}$

if I go right into differentiating, i'll get

ds/dt = t(.5)$^{t-1}$(0) so that

ds/dt = 0 , which is incorrect as we both agree.
-------------------------------------------------------

ln(s) = ln(.5$^{t}$) so that

ln(s) = t(ln(.5)) , then finally differentiating,

d/dt lns = d/dt (t(ln(.5))

d/dt lns = t'(ln(.5)) + t(ln(.5))'

d/dt lns = ln(.5) + t(1/5)(.5)'

d/dt lns = ln(.5)

(1/s)(s)' = ln(.5)

(s)' = ln(.5)(s)

(s)' = ln(.5)*(.5$^{t}$)

Which can get messy and increase likelyhood of mistakes.

------------------------------------------------------------

However, if I use the exponenent rule, i'll get

ds/dt = d/dt ln(.5$^{t}$) = ln(.5)*(.5$^{t}$)

Which is the same answer, but much more effiecient.
------------------------------------------------------------

So can it be said that when there's a constant to the power of the variable, you can't just do chain rule, or else you'll get an incorrect answer. You can technically take ln of both sides and then differentiate, but it will be messy and complicated. So the best approach would be to apply the exponental rule.

If that is correct, then my confusion now is how i would apply the exponential rule to various functions. For instance, in the original function: s(t) = (976(.835)^t - 1) +176t

Would I apply derivative rules as usual, except when I get to (.835^t), instead of using the chain rule, i use the exponent rule. Then, the 976 gets moved to the side by the constant multiple rule, the d/dx gets applied to .835^t and -1 and 176t. All the rules apply except instead of the chain rule on .835^t, the exponent rule is applied, therefore employing the use of the ln within that one term. The contant of (-1) disappears, the 176t becomes 176 and thus, the equation has been derived by using the exponent rule.

I think I understand it now, thanks!! I feel like i've made tremendous progress here in understanding this new rule of differentiation. I think i've seen the property before, but i guess it faded away since I never actually had to apply it to a problem until today. Thanks.

EDIT: Well, i at least think I have it correct and fully understood. Could you let me know if I said it correctly in t

Last edited: Mar 3, 2012
13. Mar 3, 2012

### LearninDaMath

Easy and produces the same correct answer, but requiring more work than using the exponential rule as shredder mentions. Would you agree?

14. Mar 3, 2012

### Office_Shredder

Staff Emeritus
The power rule is $\frac{d}{dt} t^n = nt^{n-1}$ A common mistake, which you appeared to try, is to do it when the variable is in the exponent: $\frac{d}{dt} a^t = t a^{t-1}$ which is not true

15. Mar 3, 2012

### LearninDaMath

Oh, thats the power rule...i just thought of it as the chain rule, but I do see the distinction. It's the chain rule, but doesn't go through the extra steps of taking the derivative of the inside term since it is automatically known that the inside term will be 1. However in this case, the derivative of the inside term will be zero, which is incorrect. I understood it, just mixed up the name for it. I used the power rule just to make the distinction between it and the exponential rule..I should have made note of that as I did in the post after that one.

P.S. I've been trying to use the "edit post" option to go back and copy & paste some of the equations i'd be using multiple times within this thread so as to avoid having to rewrite the equations over and over (still getting the hang of the latex language) and with all the copying and pasting I accidently edited the post # 1 with what I was trying to post as post #12. Luckily, I was able to piece it back together as it originally was, so there was a 2 minute span that might have been confusing if you were trying to read the first post. Anyway, thanks Shredder and Sammy for your help.

16. Mar 4, 2012

### SammyS

Staff Emeritus
Which way is more work?

It's a matter of personal preference.

Since the thread title had "log differentiation" in it, I pushed in that direction.

According to post #4, it looks like the initial problem is to differentiate the function s(t) = 976{(.835)t - 1} +176t .

After a bit of algebra:
$\displaystyle s(t)-176t+976=976(0.835)^t$​

Taking the natural log of this gives

$\displaystyle \ln\left(s(t)-176t+976\right)=\ln\left(976(0.835)^t\right)$
$\displaystyle =\ln(976)+t\ln(0.835)$​

Differentiating this gives

$\displaystyle \frac{s'(t)-176}{s(t)-176t+976}=\ln(0.835)$

Finally,

$\displaystyle s'(t) = 176 + \left(s(t)-176t+976\right)\ln(0.835)$
$\displaystyle = 176 + \left(976\left((.835)^t- 1\right) +176t-176t+976\right)\ln(0.835)$

$\displaystyle = 176 + \left(976(.835)^t \right)\ln(0.835)$

17. Mar 4, 2012

### LearninDaMath

It really is awesome to live in an age where information flows so freely ..and i'm appreciative of being able to see this problem from various perspectives. Thanks SammyS. I look at this equation now and it's not nearly as daunting as it seemed yesterday.

18. Mar 6, 2012

### LearninDaMath

post #2:

How could I use chain rule at that point if there is no composition of functions?

19. Mar 6, 2012

### Ray Vickson

You made an error going from line 4 to line 5: ln[ .835^t - 1] is NOT ln(.835^t) - ln(1).
For example, log(5-1) is not log(5) - log(1) = log(5), because 5-1 is not 5. Anyway, I honestly cannot see why you bother taking logs. If I were doing the question I would write 0.835 = e^c (c = ln(0.835), and go on from there.

RGV

20. Mar 6, 2012

### LearninDaMath

Sammy, what would be the logic or rationale for deciding to isolate 976(.835)t, or (.835)^t before taking the logarithm? ..as opposed to just taking the logarithm as it is with s(t) isolated?