quick calc question

[GoingUp]

Have an exam tomorrow and I'm a bit rusty with some of the derivatives.

I understand if y = u(x) w(x) y' = u*w' + u'*w

What I am confused about is if y = f(w) and w = g(x)

dy/dx = y'(w) * w'(x)

Can someone give me an example for the last one please? Is it a function to the power of another function?

 

[GoingUp]

bump

 

[NikPreviousAcct]

I did a semester of calc before dropping it for art, sorry

 

[orakle]

You have y = f(w) and w = g(x), and you want dy/dx of F = y * w. Let's clarify exactly what dy/dx means: differential of y over differential of x. Basically for some small movement in x you want to know how much it moves in y, or in other words the instantaneous rate of change of y with regards to x. Since f(w) does not depend on x, if i understand correctly you hold this constant (i.e. take it outside the prime) and simply take the derivative of g(x). Thus:

F = f(w) * g(x)
F' = [ f(w) * g(x) ]'
F' = f(w) * [ g(x) ]'
F' = f(w) * g'(x).

Hope that helps (I Wasn't sure exactly what you were asking).

 

[GoingUp]

Originally posted by: orakle22
You have y = f(w) and w = g(x), and you want dy/dx of F = y * w. Let's clarify exactly what dy/dx means: differential of y over differential of x. Basically for some small movement in x you want to know how much it moves in y, or in other words the instantaneous rate of change of y with regards to x. Since f(w) does not depend on x, if i understand correctly you hold this constant (i.e. take it outside the prime) and simply take the derivative of g(x). Thus:

F = f(w) * g(x)
F' = [ f(w) * g(x) ]'
F' = f(w) * [ g(x) ]'
F' = f(w) * g'(x).

Hope that helps (I Wasn't sure exactly what you were asking).

Thanks for the help but it wasn't quite what I was looking for. It looks to me like you have a function inside another function and thats what I am confused about

 

[Gibson486]

just saying that there is a function with in a function

don't worry, that stuff seems bad at first, but it is really easy....

since f is a function of w, then y will also be because it is equal to f.

All it is saying is that you can plug any function in there and derive it with respect to a certain variable.

f(w) = 2w
y = f(w)

therfore, y = 2w. See how w is the vairable? the function is now y(w)= 2w!

y' = f'(w) -> y'(w)

 

[Cooler]

f = C * g(x)
F' = y(w) * w'(x)

since y(w) is constant treat it as such

 

[Gibson486]

Originally posted by: Nik
I did a semester of calc before dropping it for art, sorry

And i thought you had strong genes....shame on you

 

[Gibson486]

Wait...I see what's going on...you are starting the chain rule....

 

[Gibson486]

Lets say we have y = (2x)^4


We could just multiply everything out, but there is another way (the chain rule).

make w = 2x

y = (w)^4

dy/dw = y' = 4w^3

noe we multiply it by w'. remember, we set w = 2x, so dw/dx = 2.

y' = y'(w) *w' = 4w^3 * 2 = 4[(2x)^3]*2 = 8(2x^3)

 

[NikPreviousAcct]

Originally posted by: Gibson486

Originally posted by: Nik
I did a semester of calc before dropping it for art, sorry

And i thought you had strong genes....shame on you

I did. Thius hot chick born in Israel with long curly black hair, sexy hips, awesome cans, and mesmerizing eyes used to give me back massages

 

[Random Variable]

In vector calculus it would be something like...

let f(u,v)=(a(u,v), b(u,v))
and let g(x,y)= (c(x,y), d(x,y))

then D(f o g) = Df(g(x,y))*Dg(x,y)

Your answer would end up being a 2x2 matrix

 

[DrPizza]

It's the chain rule that you're talking about.

When taking the derivative of a composite function, you just sort of work your way from the outermost function to the innermost function, taking one derivative at a time and leaving the rest the same... and you end up with a product.

Here's an example: f(x) = sec^5(5x^3 + 2x^2 +3x)
I would have my students re-write the trig function with the exponent on the outside, rather than the shortcut style of writing it which is generally used.

So, that means,

f(x) = [sec (5x^3 + 2x^2 +3x)]^5
The outermost function is (function ^ 5)
The next inner function is secant (another function)
Then, the innermost function is 5x^3 + 2x^2 +3x

So, first you use the exponent rule
5 * (something) ^4 times the derivative of what's inside

That something stays the same at this point, so it would be
5 * [sec(5x^3 + 2x^2 +3x)]^4 times the derivative of what's inside
Then you multiply by the derivative of the next inner function - derivative of secant.
So, it becomes 5 * [sec(5x^3 + 2x^2 +3x)] ^4 TIMES sec(5x^3 + 2x^2 +3x)tan(5x^3 + 2x^2 +3x) Times the derivative of what's inside

This then becomes 5 * [sec(5x^3 + 2x^2 +3x)] ^4 TIMES sec(5x^3 + 2x^2 +3x)tan(5x^3 + 2x^2 +3x) TIMES (15x^2 +4x)

If you wish to work inside again, the next innermost function is x. The derivative of x with respect to x (aka dx/dx) is 1

I hope it helps... it's a lot easier to teach in front of someone when you have a blackboard or a piece of paper.
<--- Friday's lesson plan: chain rule.

 

[GoingUp]

thanks to everyone for the refreshers... getting told that its a function inside another function is what I was looking for...

 

[Soccerman06]

Heres an easy one:

(5x^5) + (10x^4) + (10x^2) =

((5*5)x^(5-1)) + ((10*4)x^(4-1)) + ((10*2)x^(2-1)) =

(25x^4) + (40x^3) + (20x)

 

[iamtrout]

Originally posted by: Nik
I did a semester of calc before dropping it for art, sorry

:roll:

I guess you're just not that smart.

 

[NikPreviousAcct]

Originally posted by: iamtrout

Originally posted by: Nik
I did a semester of calc before dropping it for art, sorry

:roll:

I guess you're just not that smart.

How many other threads are you going to dig up and make some half-assed insult at one of my posts? You're starting a trend.