# Math Help - Basic Differentation Question

1. ## Basic Differentation Question

The question I'm working on is;

$d/dx[(2x+3)^3+4x^2-2/x]$

If I'm using the rules properly (a big if), I get it down to;

$3(2x+3)^2+8x+1/x$

But the answer supplied is;

$24x^2+80x+54+2x^-2$

Note: I'm not sure how to show it properly, that last part is supposed to be 2x^(-2).

Anyway, obviously, I have to continue simplifying it. I'm not sure which rules to use, and when to stop. Anyone have any tips?

2. You missed the chain rule of the first term. $\frac{d}{dx}(2x+3)^3=3(2x+3)^2*2$. The second term is correct. The last term is $2x^{-1}$ Use the power rule to see that this terms derivative is $(-1)(2)x^{-2}=-\frac{2}{x^2}$

3. That's great man, thanks a lot.

I just have 1 more general question.

I'm not sure the order in which I should/am allowed to do things. Say I have a question like this;

d/dx [(x^(-1) + 6x^(3-e) + 7x^4)/x]

Should I differentiate the top part to simplify it first, and then use the quotient rule? Or do I have to do the quotient rule first?

P.S. Is there a way to have more than one character as an exponent with the [ math ] tags? It always seems to end up like this;

$6x^3-e$ or $6x^(3-e)$

4. Originally Posted by Sucker Punch
That's great man, thanks a lot.

I just have 1 more general question.

I'm not sure the order in which I should/am allowed to do things. Say I have a question like this;

d/dx [(x^(-1) + 6x^(3-e) + 7x^4)/x]

Should I differentiate the top part to simplify it first, and then use the quotient rule? Or do I have to do the quotient rule first?

P.S. Is there a way to have more than one character as an exponent with the [ math ] tags? It always seems to end up like this;

$6x^3-e$ or $6x^(3-e)$
I think you should find the derivative in the numerator (but that is not the answer) so when you use the Quotient ring you will not have to differenciate it and get a long expression on top. Simply find the derivative and write it down somewhere. Now when you do the quotient rule you will not need to do unnecessary long computations.

5. Thanks, but now I'm getting stuck somewhere else.

What do I do with the number e?

Right now, In my answer, I end up with;

x(3-e)6x^(2-e)

and

-6x^(3-e)

That doesn't look like something that would be in a solution... What can I do with it? There's nothing in my notes or textbook that deals with it that I can find...

6. Originally Posted by Sucker Punch
Thanks, but now I'm getting stuck somewhere else.

What do I do with the number e?

Right now, In my answer, I end up with;

x(3-e)6x^(2-e)

and

-6x^(3-e)

That doesn't look like something that would be in a solution... What can I do with it? There's nothing in my notes or textbook that deals with it that I can find...
You're right it doesn't look like something that would be in a solution. But it also doesn't look like something that would be in the problem statement either. But as a matter of fact it IS in the problem statement, so you had better have terms like that in your solution.

I have a hard time legitimizing the concept of a real number exponent, but they certainly can be done. You can think of the decimal part of the number "e" as indicating a kind of radical operation, similar to the square root function being $\sqrt{x} = x^{1/2} = x^{0.5}$. A better interpretation might be out there, but I don't know of it.

By the way, left click on the LaTeX box to see how I did an exponent with more than one character. This works the same for subscripts and just about any operation that would involve more than one character.

-Dan

PS Uh oh, the left click thing doesn't seem to work any more. Okay, the rule is that if you want to express an exponent with more than one character, you write it as: x^{1/2} surrounded by the [ math ] brackets. The { } groups the characters.

7. Originally Posted by Sucker Punch
...
I'm not sure the order in which I should/am allowed to do things. Say I have a question like this;

d/dx [(x^(-1) + 6x^(3-e) + 7x^4)/x] ...
Hi,

you've got this term: $\frac{d}{dx}\left(\frac{x^{-1} + 6x^{3-e} + 7x^4}{x} \right)$

I would simplify first to avoid the quotient rule:

$\frac{d}{dx}\left(\frac{x^{-1} + 6x^{3-e} + 7x^4}{x} \right)$ = $\frac{d}{dx}\left(x^{-2} + 6x^{2-e} + 7x^3 \right)$

Consider e to be a constant real number. So you get as a result:

$(-2)x^{-3}+(2-e) \cdot 6x^{1-e}+21x^2$

EB