Question

How do I find the derivative of an equation?
Could someone give me an example? and explain it in great detail?..

Answer 1

Do you have an example to work through or just in general? Because depending on what we're doing, there are different derivative rules.

Answer 2

I know there are many forms and rules or whatever, just the basics for now , in general terms.

Answer 3

Well, you first learn the "definition" of a derivative, which is difference quotient.....but thats boring. The beginning rule is just power rule, what you do with terms that are raised to a power. In general, the power rule is this:

\[\frac{ d }{ dx }x^{n} = nx^{n-1}\]
This basically says that you take your current power and bring it down as a multiplication, then lower the power by 1. So like:

3x^2 becomes 6x
x^4 becomes 4x^3
5x becomes 5

\[\frac{ 1 }{ x } = x^{-1} \implies -x^{2} = \frac{ 1 }{ x^{2} }\] Same rule here with the fraction, just a bit trickier. This works for all powers really, square roots and the like. But the rule is the same and pretty straightforward.

Answer 4

http://tutorial.math.lamar.edu/Classes/CalcI/CalcI.aspx refer this notes for derivatives

Answer 5

Ah, I see, so
4x^2 would be 8x
x^5 would become 5x^3
and so on..

Answer 6

5x^4

Answer 7

my bad , i was looking at your example , wrote that by accident.

Answer 8

How would you go about something like \[\frac{ d^2 y }{ dx^2}\] ? I was reading on a question earlier and wondered how I would go about doing this..

Answer 9

Thats just the basic rule. But then its being able to apply it to everything accurately, even when the powers get a bit harder.

\[\frac{ 1 }{ 2 \sqrt{x}}= -\frac{ 1 }{ 4x^{3/2} }\]I think I forgot the negative on the fraction, but I cant tell, a bunch of math processing error messages, lol.

Answer 10

And there is no 'doing" that, its just notation. It means second derivative.

Answer 11

if you were to take the second derivative of something it just means you do it twice.

x^5 -> 5x^4 -> 20x^3

Answer 12

Thanks!, How would you go about applying the second derivative to equations then? Would it be the same?..

Answer 13

\[x^2 y^2 -2x =3\]

Answer 14

For example.

Answer 15

Well, solving them is the same,you just would be doing the derivative twice. Okay, now that introduces two new concepts: product rule and implicit differentiation.

Answer 16

@Psymon well explained bro :)

Answer 17

Product rule is as itsounds, when you multiply two functions/terms, etc. Theformula for that is
\[f'(x)g(x) + f(x)g'(x)\] essentially, its the derivative of the first part times the second, then plus the first times the derivative of the second. So without worrying about implicit yet, Ill do that for your x^2y^2 example. So the first part f(x) would be x^2 and the second, g(x), would be y^2.
f(x) = x^2
f'(x) = 2x
g(x) = y^2
g'(x) = 2y pluginto formula

\[2xy^{2} + 2x^{2}y\]

Oh, and the ' with the f(x) is another notation for derivative, said as prime.

y' = "y prime"
f'(x) = "f prime of x" etc. It just means 1st derivaitve.

Answer 18

Thanks :D
So what would the implicit differentiation be?

Answer 19

Well, its the method used for differentiating when you cannot easily, or at all, solve for y. Basically its the method used when you cant get y = ________

Answer 20

Doing it can be shown in different ways, but what youre actually doing is the same no matter how its notated. The idea is you pay attention to when you actually took the derivative of a y-term. So in the example x^2y^2 becoming 2xy^2 + 2yx^2, the right term, 2yx^2 has a derivative of y in it. As you differentiate everything, you keep track of all your terms with y-derivatives in them and then you eventually move all the y-derivatives to one side of the equal sign opposite of all the other derivatives. Once you have all the terms with y-derivatives opposite the other derivatives, you divide both sides of the whole equation by the entire group of y-derivatives. Now like I said, there are different ways to notate this process. I usually dont notate it much at all, but in order to show the point of all this Ill notate it and use your example above

Answer 21

and remember the derivative of a product is not the product of the derivatives.

let

Answer 22

x^2y^2 - 2x = 3 this becomes:

\[2xy^{2} + 2yx^{2}\frac{ dy }{ dx }-2 = 0\]

dy/dx represents derivative of y with respect to x. When you solve for a derivative like
y = x^2, you are technically finishing with an answer that says dy/dx = 2x. So that dy/dx is just one of a few notations for derivative of y. So now I separate my y-derivative term from everything else:

\[2yx^{2}\frac{dy}{dx} = 2-2xy^{2}\]So now when I said before the last step is todivide by all the y-derivative terms, now the reason for that becomes clear. If we take the derivative, like I said above, of something like y = 2x^2, we get dy/dx= 4x. Well, we want to solve for dy/dx, not just in these implicit problems, but period. so that's why we divide, the idea is tosolve for dy/dx. So dividing both sides by the y-derivative term I get

\[\frac{dy}{dx} = \frac{ 2(1-xy^{2}) }{ 2yx^{2} }\implies \frac{ dy }{ dx }= \frac{1-xy^{2}}{x^{2}y}\]

Answer 23

Thanks! I have an understanding of some of it now, I'll probably look into derivatives in more detail. This helped me understand a bunch, so thanks again Psymon.

Answer 24

not in calc class yet?xD

Answer 25

Nope, i'm taking pre-calc. I just want to go ahead as most of pre-calc is review of algebra 2..

Answer 26

Advance my knowledge of math.

Answer 27

Wait psymon , on the product rule \[f'(x)g(x) + f(x)g'(x)\]

I input 2x(2y) + 2x(2y)
or do you do a derivative of f(x) again because of the product rule?

Answer 28

Each term (groups separated by the plus sign) has only one derivative and that derivative is used only once. So thefirst is f derivative times g. I will never use f derivative again. Then its just f times g derivative.

Answer 29

I'm confused now ._.
f(x) = x^2
f'(x) = 2x
g(x) = y^2
g'(x) = 2y

f'(x)g(x) = 2x(2y)
right?..
so why would it be
2xy^2

Answer 30

youre multiplying derivative of f and derivative of g, but thats wrong. its derivative of f times ORIGINAL g. Then Plus. ORIGINAL F times derivative g.

Answer 31

Oh!
2x(y^2)
ah makes sense, thanks lol.

Answer 32

yeah xD there is an extension of the product rule when you have more than 2 products. Maybe when you see the pattern in that youll see the pattern ingeneral. Hereit is for 4 products, just so you can maybe pick up the pattern:

\[f'(x)g(x)h(x)i(x) + f(x)g'(x)h(x)i(x) + f(x)g(x)h'(x)i(x) + f(x)g(x)h(x)i'(x)\]

Answer 33

x^2y^2 - 2x = 3 this becomes:

2xy2+2yx2 dy/dx−2=0
I understand the 2xy^2+ part
but what happens to the 3? and the -2x?..

Answer 34

the 3 disappears and the -2x becomes -2.

Answer 35

When speaking of graphs and functions, the derivative is essentially a formula for slope. so think of this, if you have a line like y = -2x + 1. Well, that is slope-intercept form of y = mx + b, meaningthe slope is-2.

That being said, if derivative gives you slope, and I take the derivative of -2x, kinda makes sense id get -2. Because yeah, slope of -2x in a line is -2.

Now when we look at a number like 3. Well, whats the slope of 3? NOTHING. Its just the line y = 3 with a slope of 0, so taking the derivative, which finds slope, 3 just goes away because it has no slope. All constants disappear when taking the derivastive.

Answer 36

Thanks, that makes more sense now.

Answer 37

yep. So the earliest applications in calculus of taking thederivativeis to find a slope at a specific point. Because something like x^2 isnt astraight line, we normally dont have a way to say what the slope is at say x = 2. Well, the derivative gives us a way to figure that out.