Basic Calc Help

[DJBruce]

I have been trying to teach myself calculus in part by reading the tutorial. I have a few questions and was wondering is someone could give me some guidance.

 

[math]u = (x - 2)[/math]

[math]v = (x + 4)[/math]

 

The product rule says the derivative of , otherwise known as ,[imath]

(x - 2)(x + 4)[/imath], is equal to [imath]\frac{d}{dx}u \times v + \frac{d}{dx}v \cdot [/imath]. So to find the derivative of f(x), we'd do this:

[math]\frac{d}{dx} (x-2)(x+4) = \frac{d}{dx} u \cdot v = \frac{d}{dx}u \cdot v + \frac{d}{dx}v \cdot u[/math]

and then find the derivatives of the parts:

[math]\frac{d}{dx}u \cdot v + \frac{d}{dx}v \cdot u = \frac{d}{dx}(x-2) \cdot (x + 4) + \frac{d}{dx}(x+4) \cdot (x - 2)[/math]

And since we can find the derivative of things like (x - 2):

[math]1 \cdot (x + 4) + 1 \cdot (x - 2) = (x + 4) + (x - 2) = 2x + 2[/math]

 

Is there a difference when you use the [math]\cdot[/math] or [math] \times[/math]?

 

I understand where you get this:

[math]\frac{d}{dx}u \cdot v + \frac{d}{dx}v \cdot u = \frac{d}{dx}(x-2) \cdot (x + 4) + \frac{d}{dx}(x+4) \cdot (x - 2)[/math]

 

But I do not understand how you go from that to:

[math]1 \cdot (x + 4) + 1 \cdot (x - 2) = (x + 4) + (x - 2) = 2x + 2[/math]

 

I think I might understand how to do it when I do not use the chunking but I want to learn how to use the chunking method.

 

Here's what think is right.

 

[math] (x-2)(x+4)=x^{2}+2x-8 [/math]

[math]\frac{d}{dx}x^{2}=2x [/math]

[math]\frac{d}{dx}2x^{1}=1 \times 2x^{0}=1 \times 2=2 [/math]

and because [math] -8 [/math] has no variable then it is not part of my derivative

[math]\frac{d}{dx}(x-2)(x+4)=2x+2[/math]

 

Thank You any help would be greatly appreciated.

[Cap'n Refsmmat]

There's no difference between [math]\cdot[/math] and [math]\times[/math]. It's just a matter of style. Sorry for being inconsistent there.

 

Your method is correct. It's just that for more complicated problems it gets unmanageable and it becomes easier to use the product rule.

 

As for getting from this:

[math]

\frac{d}{dx}u \cdot v + \frac{d}{dx}v \cdot u = \frac{d}{dx}(x-2) \cdot (x + 4) + \frac{d}{dx}(x+4) \cdot (x - 2)

[/math]

 

to this:

 

[math]

1 \cdot (x + 4) + 1 \cdot (x - 2) = (x + 4) + (x - 2) = 2x + 2

[/math]

 

What I did was plug in the variables. We know this:

 

[math]u = (x - 2)[/math]

 

[math]v= (x + 4)[/math]

 

We can find the derivatives of each:

 

[math]\frac{d}{dx}u = 1[/math]

 

[math]\frac{d}{dx}v = 1[/math]

 

using the rules that you seem to know well. It's then a matter of inserting the variables and their derivatives into the equation from before to reach our answer.

[DJBruce]

Thank You I noticed what I was doing wrong.

 

When you said:

[math]\frac{d}{dx}u \cdot v + \frac{d}{dx}v \cdot u = \frac{d}{dx}(x-2) \cdot (x + 4) + \frac{d}{dx}(x+4) \cdot (x - 2)[/math]

I thought you ment you do:

 

[math] (v \cdot u) [/math]

 

Found the derivative of that then add it to the derivative of:

 

[math] (u \cdot v) [/math].

 

Which made no sense but now I understand that you find the derivative then multiple it.

 

Thank you for your help and thank you for your tutorial it has been a huge help and as I said before is one of the best tutorials I have seen. I am eagerly awaiting the next instalment.

[Cap'n Refsmmat]

Oh, I see. So the order of operations there isn't clear. I'll see if I can alter the post to express that in a better way.

 

Thanks for the compliments. I'll be trying to push out the next installment of the tutorial soon, as soon as I get it revised and ready to go.

[Bignose]

There's no difference between [math]\cdot[/math] and [math]\times[/math]. It's just a matter of style. Sorry for being inconsistent there.

 

In this context there is no difference between [math]\cdot[/math] and [math]\times[/math], but in higher math there becomes a significant difference.

 

These two equations have very different meanings:

 

Given vectors [math]\mathbf{a}[/math] and [math]\mathbf{b}[/math],

 

Let [math] \mathbf{c} = \mathbf{a} \times \mathbf{b} [/math]

 

and let [math] d = \mathbf{a} \cdot \mathbf{b} [/math].

 

There is a very big difference between c and d, the biggest being that c is a vector, and d is a scalar.

 

If there is a choice, Cap'n, I'd try to convert everything over to [math]\cdot[/math] or even no multiplication symbol at all (just implied), because the [math]\cdot[/math] symbol is much more consistent a wider range of mathematics.

[Cap'n Refsmmat]

I think I tried to convert everything to [math]\cdot[/math] at one point to be consistent, but I think I missed some.

[Dave]

It may be more helpful to omit any kind of dot or cross notation. When I learned the product rule back at A-level we used:

 

[math]\frac{d}{dx} (uv) = v \frac{du}{dx} + u \frac{dv}{dx}[/math]

[Bignose]

Just to kind of further was dave said here... once we got into symbolic only math, an explicit multiplication symbol was only used when it may have been ambiguous without it. I.e. if the equation was too long for a single line, there would often be an explicit multiplication symbol at the line break. But, crack open any university-level math text, or a math journal, and it is exceptionally rare to see any kind of explicit multiplication symbols. Using nothing is pretty much the standard today.

[Cap'n Refsmmat]

It may be more helpful to omit any kind of dot or cross notation. When I learned the product rule back at A-level we used:

 

[math]\frac{d}{dx} (uv) = v \frac{du}{dx} + u \frac{dv}{dx}[/math]

 

Interesting. I'd have to update the tutorial to explain what that means though.