Is the Chain Rule Needed to Differentiate 6e^{0.013t}?

[swears]

f(t) = 6e^{0.013t}

How do I find the derivative of this?

I'm confused. Do I have to use the chain rule here or the product rule, or both?

 

[EbolaPox]

You will have to use the chain rule on this. Do you know the derivative of e^x?

Example of the chain rule.
f(x) = 5(2x+1)^3. To find the derivative, use the chain rule.

dy/dx = 5 ( 3(2x +1)^2 *2). = 30(2x+1)^2.

If it helps you, you can call 2x +1 "u" and say dy/dx = dy/du * du/dx.

du/dx = 2. dy/du = 3u^2. Don't forget the constant five outside.

 

[swears]

Isen't the derivative of e^x itself? or do I need to use the natural log.

Here's my attempt from looking at your example :

6(0.013t(e)^{0.013t} * e^{0.013t}

 

[EbolaPox]

Close...

The derivative of e^u = e^u * du/dt.

So, Let's call u = .013 t

Now, we have e^u We know the derivative of e^u = e^u du/dt.

So, dy/dt = e^u * du/dt.

What is du/dt? We know u = .013t. Can we get du/dt from that? Does this help a bit?

 

[swears]

Hmm, I'm not sure if I'm getting the whole U thing sorry.

I have the chain rule being D/DX F(G(X)) = F'(G(X)) * G'(X)

I'm not even sure which part of the equation is F and which is G. If someone could explain that to me, I might have a better chance of plugging this in right.

I think it's the exponents, and the e that is throwing me off.

 

[EbolaPox]

I think you're being confused by the notation. I'll attempt to clear it up, then I"ll try to help you out.
F(g(x) means we have a function inside another function. So, let's look at what we have here:

e^.013t. That .013t is contained in an "e^x" like function, where x = .013t. Well, we can call that .013t = G(t) if we want.

Now, we want an F(g(t). We're going to call e^x "F(x)." Now, we have F(x) = e^x. . What does this mean?

Well, it means we can replace x for something like F(1) = e^1. Or, F(5) = e^5.

Or, usefully:

F(g(t)) = e^(g(t)) = e^.013t

So, Now we have that F(g(t)) form. The differentiation says

F'(g(t)) * g'(t).

This means we need two things: We need the derivative of F'(g(t)) (the derivative of e^x) and the derivative of G(t) (.013t) and then we multiply them together.

Does this help?

Example:

y = (x^2 + 3)^3 .

We want to express it in the form of F(g(x)). Notice how we have an inside function (x^2 + 3) and an outside function a Something^3. Well, we'll call the inside function g(x) = x^2 + 3. That outside ^3 function will be called F(u) = u^3

So, we have F(g(x)) = (x^2 + 3)^3 [ /tex]<br /> <br /> The chain rule says &quot;Take the derivative of the outside funciton and multiply it by the derivative of the inside funciton&quot; or F&#039;(g(x)) * g&#039;(x)<br /> <br /> So, the derivative of the outside function ^3 = 3u^2. (Remember, that &quot;u&quot; here is actually standing for our x^2 + 3)<br /> The derivative of our inside function (the x^2 +3 ) is G&#039;(x) = 2x<br /> <br /> Now, Multiply 3u^2 * 2x<br /> <br /> Remember, we said u = g(x) = x^2 +3<br /> <br /> So, our derivative is<br /> <br /> 3(x^2 +3)^2 *2x<br /> <br /> The trick here is to see an &quot;Inside function&quot; and an &quot;Outside function&quot;. THe outside function contains the inside function. Does this help at all, or do you want me to help walk you through it a bit more?

 

[swears]

Thanks for that write up.

I tried again and I got this: f&#039;(t) = .013*6e^{-.987T}(.013)

I felt like I was supposed to do something with the 6 but wasn't sure.

 

[matt grime]

the derivative of e^x is e^x, it is not e^{x-1}.

 

[EbolaPox]

Here, perhaps an example of an actual exponential function would help.

f(x) = ce^(ax^2)

Where c and a are constants.

Let u = ax^2

Let f(u) = e^u

So, the derivative dy/dx

dy/dx = dy/du * du/dx

8So, We can find du/dx = 2ax.

We can find dy/du = e^u. (The derivative of e^x is e^x, as Matt pointed out).

So, fill in where things go, dy/du = e^u du/dx = 2ax

dy/dx = 2cax * e^u Remember how I said u = ax^2? So replace u.

dy/dx = 2cax e^(ax^2)

Another example

8e^9x = y.

dy/dx = dy/du * du/dx.
u = 9x

du/dx = 9

8e^u = y. Dy/du = e^u


8e^u *9 = dy/dx = 72e^u = 72e^9x

Do these examples clear it up a bit more?

 

[swears]

Yes, thanks for your time and effort.

Now I get :

f(t) = 6e^{0.013t}

let u = 0.013t

f(t) = 6e^u

\frac {dy}{dt} = \frac {dy}{du} * \frac {du}{dt}

6e^{0.013t} * (0.013)