Print

Please login or register.
1 Hour 1 Day 1 Week 1 Month Forever
Login with username, password and session length

[Graknorke]

for f(x)=u/v
f'(x)=(vu'-uv')/(v^2)

It would be by far better to use in this case. It seems kind of weird you'd be given questions like that without being taught the quotient rule.

[Spehss _]

Meh. The point of the assigned homework is to learn the current material the class is covering. If I'm using a method to be covered in a later class to solve problems in homework in the current unit then that just seems to defeat the purpose of doing the homework in the first place.

I'm starting the whole assignment over for the third time because I'm obviously not understanding this shit, and free time is meaningless and should be spent slaving away on college work because if I don't understand this shit then I can kiss mmy grades goodbye and all my work is meaningless.


So let's say I'm to write the composite function y=cubed root(1+4x) in the form of f(g(x)) where u=g(x) and the outer function y=(f(u)).

Why couldn't I go further and say that f(u)=cubed root(u) and u(n)=1+n and n(x)=4x where y=f(u) and u=g(h) and n=h(x)? Which is right, y=f(g(x)) or y=f(g(h(x))) and how do I know which to use for the chain rule?

[MagmaMcFry]

Get the derivative of y=2/(1+e^(-x))

So y=f(g(h(x)))

f(u)=2/u

g(v)=1+v

h(w)=e^(-w)

Am I doing it right so far?

Yes.

My notes for a chain rule with 2 links is y'=f'(g(x))*g'(x). Seems like 3 links would be y'=f'(g(h(x)))*g'(x)*h'(x). Or y'=f'(g(h(x)))*g'(h'(x))

...I don't know what I'm doing.

The chain rule with 3 links is actually y'=f'(g(h(x)))*g'(h(x))*h'(x). But you don't need that. You can just use the standard chain rule on f(g(h(x))) by defining a new function j(x) = g(h(x)), then finding the derivative of j(x) by using the chain rule on g(h(x)), then finding the derivative of f(g(h(x))) by using the chain rule on f(j(x)).

[MagmaMcFry]

The point of the chain rule is to break a problem without an obvious differentiation into one that has an obvious differentiation. In your example, you can re-write the thing as (1+4x)^1/3. You want something of the form f(g(x)) because, given that, you know the derivative to be g'(x)f'(g(x)). So, you set g(x) = 1+4x, f(u) = u^(1/3). g'(x) = 4, f'(u)=u^(-2/3)/3, f'(x)=4/(3*(1+4x)^(2/3)). There's no need to break down 1+4x because you can differentiate that no problem. Sure, you could, but you would just be wasting time.

Technically, the point of the chain rule is just to break a problem down into smaller problems. The smaller problems don't have to be obvious, but you can just continue breaking them down into even smaller problems using the basic rules until they're elementary, then work your way back up from the bottom.

[Spehss _]

The chain rule with 3 links is actually y'=f'(g(h(x)))*g'(h(x))*h'(x). But you don't need that. You can just use the standard chain rule on f(g(h(x))) by defining a new function j(x) = g(h(x)), then finding the derivative of j(x) by using the chain rule on g(h(x)), then finding the derivative of f(g(h(x))) by using the chain rule on f(j(x)).

....What.

No, wait, I understand the whole thing with j(x) after rereading it a bit. That seems like unnecessary complication to me but I see why how you'd do that at least. And I guess if it works, it works.

And y'=f'(g(h(x)))*g'(h(x))*h'(x) is gotten from the whole (dy/dt)=(dy/dx)(dx/dt)=(dy/dx)(du/dx)(dx/dt). (dy/dx)=f'(g(h(x))), or the derivative of f using g as the variable, and g uses h(x) as the variable of g. (du/dx) is the derivative of g using h(x) as the variable, and (dx/dt) is the derivative of h using x as the variable. So the d(num) in the numerator specifies what the outer function and derivative is. And the d(num) in the denominator specifies...something.

What's dt supposed to be then? The derivative of x? Which is 1? So dx/dt means derivative of h(x) over 1?

[TheDarkStar]

AFAIK, the dx/dt (or dy/dx or any choice of variables) notation doesn't notate "real" division. Instead, it notates "derivative of the top variable/expression with respect to the change of the bottom variable". Although it looks like it's a fraction, it's more of a reminder of what a derivative and is also a reminder of what the formulas are (which often resemble algebraic formulas).

[ZetaX]

Unless you know what you are doing, you should indeed not treat dy/dx as if it were a normal fraction. But within the right setting it actually is a fraction; for example when using nonstandard analysis.

[bahihs]

Hey guys, I'm really stuck on this problem:

Consider the following two groups:
x* = 216            y* = 210
σ_x^2 = 2210     σ_^2 = 2618
n = 130             m = 119
where given µ_x, x_1,... ,x_n are independent and identically distributed normal
random variables with mean µ_x and variance σ_x^2
(which is known). Same for the y’s.


(d) Let us now assume that µ_x ∼ µ_y ∼ N(β,τ^2) as we did in class. Find the
posterior density f(µ_x −µ_y|x* − y*). Note this is not the same as f(µ_x −µ_y|x*, y*).
The advantage of this alternate conditioning is that it removes the dependence
on the parameter β (why?) and simplifies the posterior. Repeat part (b) for
two cases τ^2 = σ^2 and τ^2 = 2σ^2, where σ^2 = σ_x^2/n + σ_y^2/m. Hint on how to
approach this: let w = µ_x − µ_y and d = x* − y* and observe that given w, d is
normal with mean w.

Now I know the posterior density is proportional to f(d|w) * f(w) (by bayes rule), however when I do the math I get an enormous variance (something like 4σ^4) and a mean of 0. I think the problem is my assumption that f(µ_x −µ_y) ~ N(0, 2τ^2) (I'm basically assuming they have the same variance and mean, but I don't think I can and don't really know what else I can do.)

EDIT: Is there anyway to do latex on here (or something similar) the formatting is quite ugly.
Any help would really be appreciated; this hw is due tomorrow, but there is a test coming up this week and I really need to understand these concepts.

[TheBiggerFish]

That makes my calc homework look completely intelligible.
To be fair, the problem I'm having is that THE BLOODY GRAPH HAS EXACTLY ZERO BLOODY POINTS ON!
But wow.
Lots of stuff I don't know in that.

Oh.
Here's one.
Given f(x){ x^2+x, x <= 1 & 3x-2, x > 1}
Show that the function doesn't have a derivative at x = 1 using 1-sided derivatives.

[TheBiggerFish]

Oh.  One's undefined.  I had the order mixed up on which one approached from where.  Bluh.
Thanks.

[Bouchart]

Both limits are defined.  The problem is that both limits are different. (2 and 1)

[TheBiggerFish]

Wha?
No, derivatives, not limits of the functions themselves...
Although the slope at that point is infinity because of that discontinuity...

Oh.  Misunderstood the original answer.

I would go shouting Eureka but no.

[Spehss _]

Notes in calculus class today were about implicit differentiations. During notes I felt like I wasn't understanding jackshit so I tried using an example from the textbook to meticulously work through it.

Spoiler: work (click to show/hide)

"Find dy/dx of x^2+y^2=25. then find the tangent line equation at point (3,4) of the circle x^2+y^2=25"

1. So the first step is to find the differentiations of both sides of the equation.

d(25)/d(x)=0

d(x^2+y^2)/d(x) = d(x^2)/d(x)+d(y^2)/d(x)

d(x^2)/d(x)=2x

Solve for d(y^2)/d(x) using chain rule, because y is a function of x. So y^2 can be split into 2 functions in form f(g(y)) for chain rule, f(g)=g^2 and g(y)=y
f'(g)=2g
g'(y)=1

Chain rule f'(g(y))*g'(y) = 2(y)*1=2y (oh look I could've used exponential rule for that gee whiz how useful)

2. So the equation has gone from d(x^2)/d(x)+d(y^2)/d(x)=0 to 2x+2y=0 because d(x^2)/d(x)=2x and d(y^2)/d(x)=2y

3. At this point the book somehow throws in a dy/dx multiplied with 2y so the equation is currently 2x+2y*y'=0 and the initial equation somehow went from d(x^2)/d(x)+d(y^2)/d(x) to d(x^2)/d(x)+d(y^2)/d(x)*dy/dx

4. Then solve for y'
2x+2y*y'=0
2y*y'=-2x
y'=-2x/2y
y'=-x/y

b) slope=y'=-x/y
so slope at point (3,4) is -3/4

Then solve for tangent with equation y-y1=m(x-x1)
y-4=(-3/4)(x-3)
y-(16/4)=(-3/4)x+(9/4)
y=(-3/4)x-(7/4)

And that's all I've got.

Am I doing things right? And where did the dy/dx come from between steps 2 and 3 in part A? I don't give a shit how efficiently I could've done it or what other methods there are to solve the problem, I want to know if I'm doing the "implicit differentiations" shit right.

[TheBiggerFish]

Spehss_?
Please stop scaring me.


Also, the answer to my earlier question is apparently "one-sided derivatives have no bearing on the problem".  WTF, textbook writers?

[Spehss _]

You did it right.

Well that's a big relief. I feel slightly better over my chances of passing my calculus class. And working that out did help me understand the class notes better, I think.

Spoiler (click to show/hide)

Implicit differentiation just implies that you can take the derivative of y as a function of x. Remember the formula for chain rule:

dy/dx = dy/du * du/dx.

This is a rather similar thing. Set u=y

dy/dx = dy/du * du/dx
dy/dx = d/du (u^2) * du/dx
dy/dx  = 2u*du/dx
Then undo the substitution.
dy/dx = 2y*dy/dx

The way to remember it rather easily is that for any time you do an implicit differentiation of an equation involving y, for every time you take the derivative of y you have to throw in a dy/dx.

I think I understand that too. That was helpful, thanks.

Spehss?
Please stop scaring me.

My calculus class is scary, man. I used to think I was good at math.