Math Is Fun Forum

henryhightower

Guest

Derivative of Natural Logarithm

Have a problem I'm stuck on similar to:
f(x) = (x^2)(e^3x)

I don't know what rules to follow and how to go about solving this.
Is it f'(x)g(x)+f(x)g'(x)?
What does the derivative of e^3x work out to?

Another portion of the problem is to solve the f(x)=0 and I'm not sure what to do to get x by itself.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivative of Natural Logarithm

Hi henryhightower;

Is it f'(x)g(x)+f(x)g'(x)

Yes that formula is right if you want to differentiate a product of 2 functions.

What does the derivative of e^3x work out to?

Another portion of the problem is to solve the f(x)=0 and I'm not sure what to do to get x by itself.

You do not have to just plug in for x with a 0.

Please look here for explanations about functions.

http://www.mathsisfun.com/algebra/funct … ating.html

Welcome to the forum!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

henryhightower

Guest

Re: Derivative of Natural Logarithm

I wrote it wrong for the 2nd part. It's f'(x)=0.

So it'd be the derivative of x^2e^3x=0 for x.

I think that changes things.

Also, what would f'(x) look like?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivative of Natural Logarithm

Hi;

So it'd be the derivative of x^2e^3x=0 for x.

No, that is not correct. You do not need to put f(0) on one side.

In your case h(x) = x^2e^3x. So call x^2, f(x) and e^(3x) , g(x).

Can you do it now?

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

henryhightower

Guest

Re: Derivative of Natural Logarithm

d/dx(x^2)(e^3x)+(x^2)d/dx(e^3x)
so
(2x)(e^3x)+(x^2)(3e^3x) ?
not quite sure what that equals
2xe^3x+x^2(3e^3x)

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivative of Natural Logarithm

Hi;

That is correct. You have done the differentiation!

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

henryhightower

Guest

Re: Derivative of Natural Logarithm

Another portion of the problem would like me to find the values for that function equaling zero.
I believe I do that by making f'(x)=0 (the earlier reference), so in order to do so that's where I wanted to know how to isolate x to find those values.

henryhightower

Guest

Re: Derivative of Natural Logarithm

Also, appreciate the help thus far!

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivative of Natural Logarithm

Hi henryhightower;

It looks like you mean you want the roots of this, yes.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

henryhightower

Guest

Re: Derivative of Natural Logarithm

I believe so. Don't think I've really heard it referred to as the roots before. That would be the the values in which x=0?

What steps do I need to keep in mind when e(natural log, right?) is involved.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivative of Natural Logarithm

Hi;

The first thing you always do when confronted by an equation that is not a linear or quadratic polynomial. Graph the function!

Go here, I have already done it for you. Take a look and eyeball where x = 0.

http://www.mathsisfun.com/graph/functio … 2291666668

And please give katy a nudge as she is not differentiating her problem correctly.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

henryhightower

Guest

Re: Derivative of Natural Logarithm

I'm dumb. Probably from looking at so many problems today in review but it's not for x=0 so much as f'(x)=0, which I believe you did graph still. So the values where f'(x)=0 (the x-intercepts?) would be 0 and -.65?

Also, will just do some work on Katy's stuff and respond. Pretty sure they're out for the night.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivative of Natural Logarithm

0 is a root. And -.65 is an approximate answer. The proper way now to isolate the roots is through numerical methods but I will show you a boondoggle way that will doubtless make your teacher very happy.

x = - 2 / 3

Obviously only true when x = 0. So your two roots are:

x = - 2 / 3 and x = 0. There may be more on the complex plane, but these are the real ones.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.

henryhightower

Guest

Re: Derivative of Natural Logarithm

That makes sense. It's pretty much factoring them so that you can set each to = 0, right?

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: Derivative of Natural Logarithm

Hi;

Yes, works in this case. That is why you graph them first. Pays to have different ways to get an answer.

---

In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.