How Do You Differentiate a Function Involving Natural Logarithms?

[MacLaddy1]

So I've come across a derivative problem that I need to solve that is showing me some of my weaknesses in my understanding/solving of Ln and e. This is what I've done so far.

\(Q = 350\frac{1}{2}^{(\frac{t}{13.1})}\)

\(Q = 350 * \frac{1}{2}^{(\frac{t}{13.1})}\)

\(ln{Q} = \ln{350} * \ln{\frac{1}{2}^{(\frac{t}{13.1})}}\)

\(ln{Q} = \ln{350} * \frac{t}{13.1}\ln{\frac{1}{2}}\)

From here on I could keep typing, but I would rather explain. I basically took everything to the e, which eliminated the Ln wrt Q and 350, but it left me with a mess on the last term. I tried to differentiate that, but it was just a bigger mess.

Any advice, or if someone could let me know if I'm going the right direction, or a gentle (hard) kick in the correct direction, would be greatly appreciated.

Thanks,
Mac

 

[Dustinsfl]

So I've come across a derivative problem that I need to solve that is showing me some of my weaknesses in my understanding/solving of Ln and e. This is what I've done so far.

\(Q = 350\frac{1}{2}^{(\frac{t}{13.1})}\)

\(Q = 350 * \frac{1}{2}^{(\frac{t}{13.1})}\)

\(ln{Q} = \ln{350} * \ln{\frac{1}{2}^{(\frac{t}{13.1})}}\)

\(ln{Q} = \ln{350} * \frac{t}{13.1}\ln{\frac{1}{2}}\)

From here on I could keep typing, but I would rather explain. I basically took everything to the e, which eliminated the Ln wrt Q and 350, but it left me with a mess on the last term. I tried to differentiate that, but it was just a bigger mess.

Any advice, or if someone could let me know if I'm going the right direction, or a gentle (hard) kick in the correct direction, would be greatly appreciated.

Thanks,
Mac

$\ln(AB) = \ln A + \ln B\neq\ln A\ln B$

 

[MacLaddy1]

$\ln(AB) = \ln A + \ln B\neq\ln A\ln B$

Um, okay. Taking your ever so thorough post and applying it to my equation, are you saying that;

\(\ln{Q} = ln(350*\frac{1}{2}^{\frac{t}{13.1}})\)

\(\ln{Q} = ln(350) + ln(\frac{1}{2}^{\frac{t}{13.1}})\)

Can you clarify a bit?

 

[Dustinsfl]

Um, okay. Taking your ever so thorough post and applying it to my equation, are you saying that;

\(\ln{Q} = ln(350*\frac{1}{2}^{\frac{t}{13.1}})\)

\(\ln{Q} = ln(350) + ln(\frac{1}{2}^{\frac{t}{13.1}})\)

Can you clarify a bit?

Correct.

 

[MacLaddy1]

Thank you, I will work on it a bit with that correction.

---------- Post added at 08:15 PM ---------- Previous post was at 07:57 PM ----------

Okay, for an additional question regarding the last. If I take lnQ=ln(350)+ln(12t13.1), now to get rid of Ln I need to take everything to the "e," or however you say it... I'm not sure how to explain this, so this is what I mean.

\(e^{\ln{Q}} = e^{\ln{350}} + e^{\ln{(\frac{1}{2})^{\frac{t}{13.1}}}}\)

I doubt that's correct, so what's the rule here?

 

[Dustinsfl]

Thank you, I will work on it a bit with that correction.

---------- Post added at 08:15 PM ---------- Previous post was at 07:57 PM ----------

Okay, for an additional question regarding the last. If I take lnQ=ln(350)+ln(12t13.1), now to get rid of Ln I need to take everything to the "e," or however you say it... I'm not sure how to explain this, so this is what I mean.

\(e^{\ln{Q}} = e^{\ln{350}} + e^{\ln{(\frac{1}{2})^{\frac{t}{13.1}}}}\)

I doubt that's correct, so what's the rule here?

I would first use the rule $\ln A^t = t\ln A$.

Then isolate t.

Are we solving for t?

What are we doing?

 

[MacLaddy1]

We're trying to find the derivative of the original function.

 

[Dustinsfl]

From $\ln Q = \ln 350 + \frac{t}{13.1}\ln\frac{1}{2}$, take the derivative.

 

[MacLaddy1]

From $\ln Q = \ln 350 + \frac{t}{13.1}\ln\frac{1}{2}$, take the derivative.

I think this is where I have been making a mistake. I've been trying to keep Q alone to derive so I have Q', not Ln(1/Q) I'm guessing this will then turn into implicit differentiation? I'll work on this for a bit and hopefully something will click.

 

[Dustinsfl]

I think this is where I have been making a mistake. I've been trying to keep Q alone to derive so I have Q', not Ln(1/Q) I'm guessing this will then turn into implicit differentiation? I'll work on this for a bit and hopefully something will click.

Yes implicit so on the LHS you would have $\frac{1}{Q}Q'$ why will you have this on the LHS is what you need to understand.

 

[soroban]

Hello, MacLaddy!

$\text{Differentiate: }\:Q \:=\:350\cdot\left(\dfrac{1}{2}\right)^{\frac{t}{13.1}}$

I would do it like this . . .

$Q \;=\;350\cdot\left(2^{-1}\right)^{\frac{1}{13.1}t} \;=\;350\cdot(2)^{-\frac{1}{13.1}t} $Take logs: .$\ln Q \;=\;\ln\left(350\cdot(2)^{-\frac{1}{13.1}t}\right) \;=\; \ln(350) - \frac{1}{13.1}t\ln(2) $Differentiate implicitly:

. . $\dfrac{1}{Q}Q' \;=\;0 - \frac{1}{13.1}\cdot1\cdot\ln(2) \;=\;-\dfrac{\ln(2)}{13.1} $

. . $Q' \;=\;-\dfrac{\ln(2)}{13.1}Q \;=\;-\dfrac{\ln(2)}{13.1}\cdot350\left(\frac{1}{2} \right)^{\frac{t}{13.1}} $

. . $Q' \;=\;-\dfrac{350\ln(2)}{13.1}\left(\dfrac{1}{2}\right)^{\frac{t}{13.1}}$

 

[MacLaddy1]

Thanks dwsmith and soroban,

I had just about worked my way through it when soroban posted, and I saw his shortcut. Is that a good, common, way of doing things with differentiation? If I see $\frac{1}{x}$ should I treat it as $x^{-1}$. It's something I've seen before, of course, but it hasn't been taught as something I should really look out for.

Again, I appreciate both of your help.

---------- Post added at 10:15 PM ---------- Previous post was at 10:02 PM ----------

Additionally, playing with these numbers more, I just realized what I'm sure is/was a large gap in my Logarithm understanding. I didn't realize that $ln(1/x) = -ln(x)$

That sure would have saved some time... Oh well, hope I don't have too many more large holes in my head.

 

[Dustinsfl]

Thanks dwsmith and soroban,

I had just about worked my way through it when soroban posted, and I saw his shortcut. Is that a good, common, way of doing things with differentiation? If I see $\frac{1}{x}$ should I treat it as $x^{-1}$. It's something I've seen before, of course, but it hasn't been taught as something I should really look out for.

Again, I appreciate both of your help.

---------- Post added at 10:15 PM ---------- Previous post was at 10:02 PM ----------

Additionally, playing with these numbers more, I just realized what I'm sure is/was a large gap in my Logarithm understanding. I didn't realize that $ln(1/x) = -ln(x)$

That sure would have saved some time... Oh well, hope I don't have too many more large holes in my head.

Just like we had $\ln (AB) = \ln A + \ln B$, there is also a division rule $\ln (A/B) = \ln A - \ln B$.

For $\ln (1/x) = \underbrace{\ln 1}_{0} - \ln x$.

Just applications of the rules. Once you have them down, it will all be easy.

 

[MacLaddy1]

$\ln (1/x) = \underbrace{\ln 1}_{0} - \ln x$

Ah, that makes much more sense. Thank you for pointing that out.

 

[CaptainBlack]

So I've come across a derivative problem that I need to solve that is showing me some of my weaknesses in my understanding/solving of Ln and e. This is what I've done so far.

\(Q = 350\frac{1}{2}^{(\frac{t}{13.1})}\)

\(Q = 350 * \frac{1}{2}^{(\frac{t}{13.1})}\)

\(ln{Q} = \ln{350} * \ln{\frac{1}{2}^{(\frac{t}{13.1})}}\)

\(ln{Q} = \ln{350} * \frac{t}{13.1}\ln{\frac{1}{2}}\)

From here on I could keep typing, but I would rather explain. I basically took everything to the e, which eliminated the Ln wrt Q and 350, but it left me with a mess on the last term. I tried to differentiate that, but it was just a bigger mess.

Any advice, or if someone could let me know if I'm going the right direction, or a gentle (hard) kick in the correct direction, would be greatly appreciated.

Thanks,
Mac

\[Q(t)=350\times (1/2)^{t/13.1}\]

Now \(x=e^{\ln(x)}\) by definition, so:

\[Q(t)=350\times e^{[\ln(1/2) \times t]/13.1}=350\times e^{[\ln(1/2)/13.1]\; t}\]

etc...

CB

 

[MacLaddy1]

\[Q(t)=350\times (1/2)^{t/13.1}\]

Now \(x=e^{\ln(x)}\) by definition, so:

\[Q(t)=350\times e^{[\ln(1/2) \times t]/13.1}\]

etc...

CB

I'm glad you brought that up, Captain, because it raises another question that I had. (even though I'm not sure about your exponent. Why would you multiply ln(1/2) by t just to divide by 13.1)
Anyway, back to the question.

Since $x = e^{\ln(x)}$, and I see you can "e" the one term (or whatever the verb is), then don't you have to "e" everything in the equation, or can you just do it to the one term?

 

[CaptainBlack]

I'm glad you brought that up, Captain, because it raises another question that I had. (even though I'm not sure about your exponent. Why would you multiply ln(1/2) by t just to divide by 13.1)
Anyway, back to the question.

Since $x = e^{\ln(x)}$, and I see you can "e" the one term (or whatever the verb is), then don't you have to "e" everything in the equation, or can you just do it to the one term?

You use this to convert something raised to a power of a function of the variable to "e" raised to some function of the variable so that you can use the rules for differentiating the exponential function rather than having to remember the appropriate rule for an arbitrary.

The alternative is to use the rule:

\[ \frac{d}{dt} k^{at} = a \ln(k) k^{at} \]

CB