Homework Help: Having trouble differentiating exponential equations

1. Apr 24, 2009

Draggu

1. The problem statement, all variables and given/known data
L(t)=15(0.5^(t/26))
Find the rate of L, when t=60

2. Relevant equations

3. The attempt at a solution

L'(t) = (15/26)(1/2)^(t/26)ln(1/2)
L'(60)=(15/26)(1/2)^(60/26)ln(1/2)

= -0.08

Did I do this right? If I did it wrong, please say where, I am having great trouble understanding this.

2. Apr 24, 2009

Cyosis

Yes you did it correctly. You say you have great trouble understanding "this". What is "this" exactly, how to differentiate an exponent?

3. Apr 24, 2009

Draggu

I guess I was just lucky to be honest, since I did the question with a friend. My problem is applying the chain rule, and knowing where/when to put ln.

Here's another, that I have not finished yet (don't know how)

Flow of lava from a volcano is modelled by

l(t)= 12(2-0.8^t)

l is dist from crater in km
t is time in hours

how fast is the lava travelling down hillside after 4hours,

well, firstly i need to differentiate l(t), then solve for t

Can somebody give me hints for what to do with the t? I'm not quite sure..

4. Apr 24, 2009

Dick

0.8^t=e^(ln(0.8)*t). The derivative of e^(ct) with respect to t is what? Use the chain rule. Powers of constants are just exponentials.

5. Apr 24, 2009

jhae2.718

Use this derivative shortcut for exponential functions:
dy/dx a^u=ln(a)*a^u*du //*du is the derivative of u; this is where you use the chain rule
For a=e, dy/dx e^u=e^u*du

You are supposed to find the rate at which lava flows down the hill at 4 hours. Since t is the time in hours and l(t) is the distance of the lava from the crater in km, you need
to find l'(4) to get the rate. It is not necessary to "solve" for t: t=4.

If you were told the rate and were supposed to find the time, t, you would use algebra to solve for t, much as any other equation. Since l'(t) is an exponential function, you would have to use logarithms and the property that log(a^b)=b log(a), or that log(a)/log(b)=log_b(a).

6. Apr 25, 2009

HallsofIvy

An exponential is about the easiest function to differentiate!

(ex)'= ex and (ax)'= ax ln(a).

More generally, by the chain rule, (af(x))'= af(x)ln(a) f'(x).

7. Apr 25, 2009

Draggu

l(t)= 12(2-0.8^t)

so l'(t) = 12(2-0.8^(t) ln 0.8)

If that's true, it doesn't seem to get me anywhere.

8. Apr 25, 2009

Hootenanny

Staff Emeritus
That's not quite right, why is there still a 2 there?

9. Apr 25, 2009

Draggu

Ah yes.

so l'(t) = 12[-0.8^(t)] ln 0.8)

..
..

So when I sub in 4 for t, I get 1.1. Is this correct?

10. Apr 25, 2009

Hootenanny

Staff Emeritus
Indeed you do.

11. Apr 25, 2009

Draggu

"how fast is the lava travelling down hillside after 4hours"

One more question, just needs clarifying.

For example if I had (1/2)^(t/138)

The derivative would be 1/138(1/2)^t/138 ln 1/2 ?

I'm having trouble differentiating an exponent with a denominator

12. Apr 25, 2009

Hootenanny

Staff Emeritus
Apologies, I was looking at your OP.

13. Apr 25, 2009

jhae2.718

That derivative would be correct. With a denominator, remember the chain rule" dy/dx=dy/dt*dt/dx.

To put it simply, you multiply the derivative of the function (the a^x*ln(x) ) by the derivative of the exponent.

So for a derivative with a denominator, you would have a^(x/b)*ln(a)*dx/b.

Examples: dy/dx 2^(x/20)=2^(x/20)*ln(2)*1/20
dy/dx 3.4^[(x^2)/85]=3.4^[(x^2)/85]*ln(3.4)*(2x)/85
dy/dx 3-5^[(x^4)/220]=-5^[(x^4)/220]*ln(5)*[(x^3)/55]

Does that help?