/sci/ - Science & Math

[Anonymous No.3072646]

So /sci/, can you teach a retard to find the derivative of a function? I challenge you to teach me how to. I'm in college algebra now.

[Anonymous No.3072651]

consult the sticky

[Anonymous No.3072650]

in general or just a specific function?

[Anonymous No.3072653]

1. go to http://www.wolframalpha.com/
2. type "differentiate [your function]"

Done

[Anonymous No.3072652]

the limit as h approaches 0 of [f(x-h)-f(x)]/h

>ITT i don't know laytex

[Anonymous No.3072654]

>>3072650
In general. Just the general rules to follow for finding the derivative of any function. I'm seriously an idiot, I had an 83% in college algebra, but I just like playing with math.

[Anonymous No.3072655]

what the hell is college algebra?

[Anonymous No.3072661]

>>3072652
So its simply the limit of the difference quotient?

[Anonymous No.3072662]

>>3072655

The only college algebra I took was abstract algebra.

[Anonymous No.3072664]

>>3072655
basic algebra. exponential functions, logarithms, etc.

[Anonymous No.3072665]

Take the power down and multiply the co-efficient. Take one away from the power.

You're welcome.

[Anonymous No.3072669]

>>3072665
*multiply the co-efficient by the power

derp

[Anonymous No.3072675]

>>3072661

Yea, its just the slope of a tangent line, so as h -> 0

[Anonymous No.3072673]

>>3072661
Yes, basically.

[Anonymous No.3072679]

>>3072661
not quite the numerator is f(x-h) - f(x) and the denominator is simply h. You need to take the limit as you can not divide by zero, the limit is the derivative... There are also simple rules, for those see the sticky like everyone else here says

[Anonymous No.3072685]

find f(x+h) - f(x)
divide everything by h
if you still see h anywhere, draw a large X through those terms
hooray

alternatively use various shortcuts

[Anonymous No.3072687]

>>3072673
Well goddamn that's fucking easy. And to think I was worried about taking calculus...So is there any other things I should know about? I've heard terms like "chain rule" and "product rule" before, what do they mean?

[Anonymous No.3072699]

>>3072687
just procedures for differentiating different types of expressions. differentiation isn't hard, but it can get tedious if the expressions are complex enough. college algebra was harder than calc 1 for me.

[Anonymous No.3072700]

Thanks everyone. That's pretty easy. So what is an integral?

[Anonymous No.3072717]

>>3072700
The antiderivative of a function basically. Also signifies the area under the graph of the function, where the derivate is the slope of the graph at a point.

If you derivate something and then integrate it you will end up at the starting point.

[Anonymous No.3072723]

>>3072717
So is there a basic equation I can also use to take the integral of something, like there is with the derivative? Also, what does the fancy notation of an integral mean (the one that looks like a stretched "s")?

[Anonymous No.3072738]

>>3072687
what I provided there was the defention of a dervitive, the rate in change in a curve... if you actually intrested still take a course so you will learn the chain rule, product rule, quotient rule, implicit differentiation and so much more including integrals and such... I'm a physics major and I have to take 2.5 years of calc... there is so much more then a simple limit...

[Anonymous No.3072746]

If you just want to know the basic of it then it's just remembering a bunch of rules and then you just gotta apply those rules... f(x)=2x^3-4+4x would differentiate into f '(x)=6x^2+4. the reason 2x^3 differentiates into 6x^2 is because you multiply the number (2) with the number equivalant to the power, like 2*3x^3 but the you also gotta substract 1 from the power so that 2*3x^3-1 and therefore 6x^2. The reason the 4 just dissapears is because whenever there is a single number (doesn't matter if it's negative or positive) it just equals 0. The last part of the function, 4x, differentiates into 4 because, in a way, you can just say that when the x is 1 and it multiplies the number (this isn't exactly how it happens but nevermind that.) There is also a lot of other rules like f(x)=cos(x) differentiates into f '(x)=-sin(x) and such.

http://www.dummies.com/how-to/content/the-basic-differentiation-rules.html

[Anonymous No.3072752]

>>3072738
I'm definitely interested. I'm only taking these classes because I like the subject matter (even though I suck at it). I'm enrolled in Calculus 1 next semester actually.

[Anonymous No.3072761]

>>3072723
when you have some easy function, add 1 to the power, and divide the coefficient by the new power
cx^n => (c/n+1)x^(n+1)
for example 2x^2 => (2/3)x^3 (try to differentiate it and you will have the original function)

fancy notation means that you will be antidifferentiating a given function. if there are some numbers on top and below it, these are limits and they're used when you need an area.

[Anonymous No.3072756]

>>3072665

f(x)=x^x
f' '(x)=/= x(x)^(x-1)

your definition is shit and at best only applicable in certain instances.

[Anonymous No.3072767]

>>3072746
So is a the reason a derivative of a single number (say, 4) is zero because the graph of 4 would be a straight line?
And I definitely understand what you did with those numbers, multiplied the two by the power and reduced the power by one.

[Anonymous No.3072771]

>>3072756
yup, and this, op, is when you use a chain rule - function in a function

[Anonymous No.3072796]

>>3072771

You knew this shit was coming

[Anonymous No.3072802]

>>3072771
So how would one go about differentiating f(x) = x^x? because my initial thought was the differentiation would be x(x)^x-1 going by the method mentioned above, where you multiple the front by the power and reduce the power.

Also, if I seem to disappear for a bit, I'm taking my girlfriend to lunch for 30 minutes at 12:30 and I should be back by 1:30. I appreciate the help everyone :)

[Anonymous No.3072811]

>>3072767

Yes.

[Anonymous No.3072826]

>>3072802

logarithmic differentiation

[Anonymous No.3072845]

>>3072802
Rewrite it as e^(ln(x^x))
Now apply the rule that ln(a^b) = b*ln(a)
Get e^(x*ln(x))
Now differentiate normally

[Anonymous No.3072848]

<span class="math">
f(x) = x^y
f'(x) = yx^(y-1)
[/spoiler]

[Anonymous No.3072860]

Powerule
f(x)=x^(n) ⇔ f '(x)=n*x^(n-1)
f(x)=x^(-n) ⇔ f '(x)=-n*x^(-n-1) or alternatively f(x)=x^(-n) ⇔ f '(x)=-n*x^(-(n+1))
If you fx. have the function f(x)=2x^(3) then it differentiates into f '(x)=6x^(2) because f '(x)=2*3x^(3-1).

Constant rule
f(x)=c ⇔ f '(x)=0
If you fx. have the function f(x)=5 then it differentiates into f '(x)=0 bcause f(x)=5 is just a straight line.

Sum rule
If there are several terms in your function then just differentiate every single one.
f.eks f(x)=ax^(n)+2x+c ⇔ f '(x)=a*nx^(n-1)+2

Differentiating a fraction.
If you fx differentiate. ((f(x))/(g(x))) then you make a new fraction ((g(x)f '(x)-g '(x)f(x))/((g(x))^(2))).

Differentiating a multiplied expression.
If you fx. differentiate (f(x))*(g(x)) then you make a new expression called (g '(x))*(f(x))-(g(x))*(f '(x)).

Differentiating of trigonometric functions and other miscellaneous functions.

f(x)=sin(x)⇔f '(x)=-cos(x)
f(x)=cos(x)⇔f '(x)=sinx
f(x)=tan(x)⇔f '(x)=sec^(2)(x)
f(x)=sec(x) f '(x)=sec(x)tan(x)
f(x)=cot(x) f '(x)=csc^(2)(x)
f(x)=csc(x) f '(x)=csc(x)cot(x)
f(x)=ln(x)⇔f'(x)=((1)/(x))
f(x)=e^(x)⇔f'(x)=e^(x)
f(x)=log(x) f '(x)=((log(e))/(x))
f(x)=a^(x) f '(x)=a^(x)*ln(a)

[Anonymous No.3072861]

>>3072802
<span class="math">y = x^x[/spoiler]
<span class="math">ln y = ln (x^x) = x ln x[/spoiler]
<span class="math">\frac{y'}{y} = ln x + x * \frac{1}{x}[/spoiler]
<span class="math">y' = y(ln x + 1) = x^x(ln x + 1)[/spoiler]

[Anonymous No.3072856]

>>3072845
Oh that makes sense! Well hell, this stuff really doesn't seem terribly hard. You just have to play with the equation until it is in a form that you can use to differentiate normally.

I'm heading out now and will be back in around an hour. Thank you so much! I only have a few more questions when I get back.

[Anonymous No.3072889]

For me, derivatives were some of the coolest shit I learned in math. I had a blast doing the beginning portions of calculus.

[Anonymous No.3072927]

>>3072889
this

[Anonymous No.3073088]

>>3072860
OP back from lunch. That is fantastic, thank you! So in regards to physics, how is derivatives used?

[Anonymous No.3073173]

>>3073088

"A derivative is a rate of change, which, geometrically, is the slope of a graph. In physics, velocity is the rate of change of position, so mathematically velocity is the derivative of position. Acceleration is the rate of change of velocity, so acceleration is the derivative of velocity. Net force is the rate of change of momentum, so the derivative of an object's momentum tells you the net force on the object. These are only a few of the applications of the derivative in physics."

[Anonymous No.3073191]

>>3073088

Good question.

Let's say we have a function that models the distance that an object travels with respect to time, so D(t).

If we differentiate D(t), we're basically deriving a function that gives us the change in distance with respect to time, which is velocity so the derivative of D(t) [or D'(t)]is V(t).

If we differentiate V(t), we derive a function that gives us the change in velocity with respect to time, or acceleration. So in other words the derivative of V(t) [which is V '(t)] is A(t)

TL;DR

The derivative of a position function is velocity, the derivative of a velocity function is acceleration etc etc.

[Anonymous No.3073204]

y = x^x = e^[x*ln|x|]
chain rule
dy/dx = e^[x*ln|x|]*(1*ln|x|+x/x) = x^x[ln|x|+1]

[Anonymous No.3073207]

I did >>3073204
But I like
>>3072861
better.

[Anonymous No.3073224]

>>3073088

Btw, it's refreshing to see someone excited about math =). Been awhile since i've seen such curiousity

[Anonymous No.3073260]

>>3073088
Derivatives are used mostly for rates of change.
Velocity is the derivative of position in terms of time, current is the derivative of charge in terms of time, emf is the derivative of magnetic flux in terms of time, etc.