Having trouble differentiating exponential equations

[Draggu]

Homework Statement

L(t)=15(0.5^(t/26))
Find the rate of L, when t=60

Homework Equations

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.

 

[Cyosis]

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

 

[Draggu]

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

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 modeled 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 traveling 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..

 

[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.

 

[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).

 

[HallsofIvy]

An exponential is about the easiest function to differentiate!

(e^x)'= e^x and (a^x)'= a^x ln(a).

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

 

[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.

 

[Hootenanny]

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.

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

 

[Draggu]

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

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?

 

[Hootenanny]

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?

Indeed you do.

 

[Draggu]

Because the question asked
"how fast is the lava traveling 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

 

[Hootenanny]

Because the question asked
"how fast is the lava traveling down hillside after 4hours"

Apologies, I was looking at your OP.

 

[jhae2.718]

Because the question asked
"how fast is the lava traveling 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

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?