Question

can you show me step by step and what rule do I use in this calculus problem and why? using d/dx (1 + x)/ |/x\|)=

Answer 1

\( \displaystyle \frac{d}{dx} \frac{1 + x}{\left| x \right| } \)
Or am I mistaken at what the denominator is supposed to be?

Answer 2

The x in the denominator is supposed to be square root and the entire equation has parenthesis around it.

Answer 3

Oh, got it.
\( \displaystyle \frac{d}{dx} \left( \frac{1 - x}{\sqrt{x}}\right) \)
Like this then? This looks a lot better. (Absolute values aren't fun to play with).

Answer 4

Yes

Answer 5

The first thing to do here is using the quotient rule. We would use this because we have a distinct fraction of two functions.

Answer 6

Are you familiar with quotient rule for derivatives?

Answer 7

No

Answer 8

We have been going over so many rules and formulas in a short period of time and its hard to remember all of them or even when to use them.

Answer 9

Alright. Does quotient rule at least ring a bell as something covered? I could write that one out.

Otherwise, we can use product rule with a bit of manipulation.

In general, a rule is named after what it is used for. Like...
Power Rule: when you have powers. d/dx (x^ n) = n x^ (n-1)
Product Rule: when you have a product: d/dx (f(x) g(x)) = d/dx f(x) g(x) + f(x) d/dx g(x)
etc.
But overall, memorization is the best thing for the rules.

Answer 10

Yes, she did cover the quotient rule, but very quickly.

Answer 11

I would say it is best to memorize all of the rules. I recently just got the idea that flashcards could be awesome for studying the rules you have and their usage.

But, anyways, here is the quotient rule:
\( \displaystyle \frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{ \frac{d}{dx} (f(x)) g(x) - f(x) \frac{d}{dx} (g(x))}{(g(x))^2} \)

Answer 12

We would just call f(x) = 1 - x, and g(x) = sqrt(x)

Answer 13

1 + x, not 1-x

Answer 14

Yes, it is 1 + x

Answer 15

We would just directly apply the two functions.
\( \displaystyle \frac{d}{dx} \left( \frac{1 + x}{\sqrt{x}} \right) = \frac{ \frac{d}{dx} \left(1 + x \right) * \sqrt{x} - (1 + x) * \frac{d}{dx} \left(\sqrt{x} \right)}{(\sqrt(x))^2} \)

Answer 16

ok, so when using the rule you would just plug it in?

Answer 17

Yeah, most rules are expressed as a very general case. The quotient rule is with two general functions f(x) and g(x). If we have a fraction f(x) / g(x), we just set the two functions as the numerator and denominator and make the substitution into the statement of the rule.

Answer 18

Usually, there will be a bit more to do such as this case after we substitute everything. Note that we have two derivatives in the new expression: d/dx (1 + x) and d/dx( sqrt(x) ).

Answer 19

But they are much easier to deal with than the original derivative!
d/dx(1 + x) is very easy, just using the rules for addition and constants.
d/dx(sqrt(x)) is a tad harder. You have to rewrite sqrt(x) = x^ (1/2) as a rational power. This is now a power rule derivative.

Answer 20

So, why would sqrt(x)=^(1/2)?

Answer 21

It is more or less a matter of conventions. But if we can agree that:
\( \sqrt{x} ^ 2 = x \) (Well, ignoring all the unusual square -and-square root stuff)
And
\( \left( x^{1/2} \right)^ 2 = x^{ 2 * 1/2} = x \)
The comparison might make more sense.

Answer 22

So, do you use this all the time when dealing with a situation like this?

Answer 23

Whenever you have a radical, you will pretty much always convert it into a rational power. Yes.

Say you have: \( \sqrt[3]{x} \), You would rewrite it as \(x^ {1/3} \).

Answer 24

And then from there, power rule always applies like this:
\( \displaystyle \frac{d}{dx} \left( x^ n \right) = n x^{n-1} \)

If n is 1/2
\( \displaystyle \frac{d}{dx} \left(x^{1/2} \right) = \frac{1}{2} x^ {1/2 - 1} \)

Answer 25

Does it matter if its in the numerator or denominator?

Answer 26

It should not matter where it is in the function which you are taking the derivative of. If you really had to, you could move it around by taking it to a negative power. But in the end, as long as the rules are applied correctly it won't change the answer in the end.

Answer 27

Like, if you had this:
\( \displaystyle \frac{d}{dx} \left( \frac{1}{\sqrt{x}} \right) \)
You can rewrite the radical...
\( \displaystyle \frac{d}{dx} \left( \frac{1}{x^{1/2}} \right) \)
If you used quotient rule here, you would get the right answer. However, you can move it to the numerator by making it negative power:
\( \displaystyle \frac{d}{dx} \left( x^{-1/2} \right) \)
This one is now able to be approached strictly with power rule.

A sense of which way is preferable mainly comes with practice.

Answer 28

The sense for finding any derivative comes with experience, though. It may seem contrived to use stuff like quotient rule at first but if you use it enough times, it can become second-nature.

Answer 29

Ok, I know I'm asking a lot of you do you think you could tutor me?

Answer 30

I am happy to help when I can/am online, but my schedule here is often sketchy at best. Plus there are many other fine helpers on OS! :)

Answer 31

Ok, well I do appreciate any help that I can receive. Especially when one is explaining it in the simplest form for me to understand.

Answer 32

Ok, so after I have applied the two functions where would I go from there?

Answer 33

From
\( \displaystyle \frac{d}{dx} \left( \frac{1 + x}{\sqrt{x}} \right) = \frac{ \frac{d}{dx} (1 + x) \sqrt{x} - (1 + x) \frac{d}{dx} \left( \sqrt{x} \right) } { (\sqrt{x} )^2 } \)

We would now need to find those two derivatives in the numerator on the right-side.
We can do them individually and substitute the result.

Answer 34

\( \displaystyle \frac{d}{dx} \left( 1 + x \right) \)
This one is the easier of the two. The notable rules to use would be Addition Rule (or Linearity) and then Constant/Power Rules.

Answer 35

Addition Rule: d/dx (1) + d/dx (x)
Constant Rule: d/dx(1) = 0 As for the derivative of any constants.
Power Rule: d/dx(x) = d/dx ( x^ 1) = 1 * x^ (1-1) = 1*x^0 = 1.

Answer 36

So this means \( \displaystyle \frac{d}{dx} \left( 1 + x \right) = 1\), putting it all together. That one is clear?

Answer 37

Yes

Answer 38

And then we have the derivative of the sqrt(x) function, which I already mentioned but it never hurts to see it all together again.
\( \frac{d}{dx} \left( \sqrt{x} \right) = \frac{d}{dx} \left( x^ {1/2} \right) \)
Power Rule: d/dx (x^ (1/2) ) = 1/2 * x^ (1/2 - 1) = 1/2 x^ (-1/2)
We can move x^(-1/2) to the denominator, then, making the power positive. Put it back as a sqrt(x) instead of x^ (1/2) and we are left with:
\( \displaystyle \frac{d}{dx} \left( \sqrt{x} \right) = \frac{1}{2 \sqrt{x}} \)
This is clear as well, then? :)

Answer 39

If those are both clear, then the last step is just replacing them in the original equation.
\( \displaystyle \frac{d}{dx} \left( \frac{1 + x}{\sqrt{x}} \right) = \frac{\color{green}{\frac{d}{dx} (1 + x)} \sqrt{x} - (1 + x) \color{green}{\frac{d}{dx} (\sqrt{x}) }}{x} \)

Answer 40

It's a little clear. I know that I will have to constantly practice, but your making helping me to understand.

Answer 41

Yeah, if it ever seems unclear, I can try my best to give you a better idea.

Inevitably, I can't blame you for being at least somewhat uncertain! Heck, my friends in AP Calc are into integration and still forget what the quotient rule even is. lol

Answer 42

lol

Answer 43

Well at least I know I'm not by myself!

Answer 44

A bit of a summary, then;
1.) We used Quotient Rule on the original function. d/dx(f / g) = [ df(x)/dx g(x) - f(x) dg(x)/dx ] / [ g(x) ]^2. The cue is just that we have a function in the numerator and denominator.
2.) We used Addition Rule to simplify d/dx(1 + x) and then constant/power rules to find d/dx(1 + x) = 0 + 1 = 1.
3.) We rewrote sqrt(x) as x^(1/2) to apply power rule. d/dx (x^(1/2)) = 1/2 x^(-1/2).
4.) Resubstitute the derivatives and find the original answer!

Answer 45

I'd say, in terms of difficulty, quotient rule is the most difficult to memorize. A close second would be the chain rule. Product rule next, then power rule, constant rules, and addition / subtraction etc. sort of come naturally.

If you become fluid in applying at least the product rule, power rule and chain rule, you will be golden for most derivatives not involving special functions like trig functions.

Answer 46

I will make flash cards like you advised me to do. I know that it will help tremendously. I have a wonderful Instructor, but she cannot talk fluid English and when you ask if she could go back her reply is don't worry everything will be OK.

Answer 47

Yes, and you will be able to incorporate more of the special functions as well when the time comes (hopefully they are not all thrown at once)!
It may also help to go over sections ahead of time, just like skimming over what is to come. Having a bit of mind on what is happening next could probably make it piece together more cleanly. :)

Answer 48

Anyways, are there any other questions before I go off for the night? Getting late here (10:30 pm). :)

Answer 49

Yes, so, once I apply all the steps above how will my problem look?

Answer 50

You should end up with something like this:
\( \displaystyle \frac{d}{dx} \left( \frac{1 + x}{\sqrt{x}} \right) = \frac{\sqrt{x} - \frac{x + 1}{2\sqrt{x}}} { x} \)

Oftentimes, you need not simplify something into a simplest form unless (a) the teacher wants you to do so always or (b) it asks you to do so. It is usually sufficient to simply have all of the derivatives taken.

Answer 51

Ok, I really appreciate your time. Thanks for helping. Have a good night.

Answer 52

You're welcome! Best of luck in your studies and good night! :)

Answer 53

Good night.