help..the chain rule

~Candy~

this is the q
3sin^5(x^2-3x)

usually what we do with chain rule is

say (2x+1)^3
=3(2x+1)^2*2
put 3 up in the frond then rewrite the (2x+1)^(3-1)
like i know where is the *2 come from , you just diff the inside bit..


well this question i got as far as
15(sin(x^2-3x)^4 * ( i don't get this part) the answer says
(cos(x^2-3x))*(2x-3)

I was just wondering where did the cos(x^2-3x part come from..
I just couldn't get it =(( ...

ehh, confusing..did i make myself clear?
i scanned the question..

thanks for your help!!!

Thomas_S_Hunterson

easy way to do it is let
u = x^2 - 3x
v = sin(u)
y = 3v^5

This is breaking the function down into easily differentiable parts. Differentiate all these to get du/dx, dv/du, and dy/dv respectivelly.

and then
dy/dx = (du/dx)*(dv/du)*(dy/dv)

1huge1

Honours calculus I assume, Glad I dropped before we got to that.

In pass the chain rule never has anything like Sin in it.

Davidius

When doing chain rule you just need to multiply by d/dx whatever is in the bracket. (sin(x^2-3x))

I believe that comes from chain rule again. If u is (x^2-3x) and a = sin(x^2-3x) or a = sin u then da/dx = da/du*du/dx

da/du = cos u ; du/dx = 2x-3

Meaning da/dx = cos u * (2x-3) = cos(x^2-3x)*(2x-3)
[d(sin angle)/dx = cos angle according to log tables]

da/dx = cos(x^2-3x)*(2x-3)

With a = sin(x^2-3x) then y = 3a^5

dy/dx = dy/da*da/dx (chain rule)

dy/da = 15a^4

dy/dx = 15a^4 * cos(x^2-3x) * (2x-3)

dy/dx = 15(sin(x^2-3x))^4 * cos(x^2-3x) * (2x-3)

JC 2K3

Well, there are 3 things going on here; a sine of an polynomial to the power of 5. So all you do is differentiate each of them and multiply them together (along with the constant(3)).

I've illustrated this with nice colours

3sin^5(x^2-3x) = 3*((sin(x^2-3x))^5)

3*(5(sin(x^2-3x)^4)*cos(x^2-3x)*(2x-3)

~Candy~

Sean_K wrote: »

easy way to do it is let
u = x^2 - 3x
v = sin(u)
y = 3v^5

This is breaking the function down into easily differentiable parts. Differentiate all these to get du/dx, dv/du, and dy/dv respectivelly.

and then
dy/dx = (du/dx)*(dv/du)*(dy/dv)

Davidius wrote: »

When doing chain rule you just need to multiply by d/dx whatever is in the bracket. (sin(x^2-3x))

I believe that comes from chain rule again. If u is (x^2-3x) and a = sin(x^2-3x) or a = sin u then da/dx = da/du*du/dx

da/du = cos u ; du/dx = 2x-3

Meaning da/dx = cos u * (2x-3) = cos(x^2-3x)*(2x-3)
[d(sin angle)/dx = cos angle according to log tables]

da/dx = cos(x^2-3x)*(2x-3)

With a = sin(x^2-3x) then y = 3a^5

dy/dx = dy/da*da/dx (chain rule)

dy/da = 15a^4

dy/dx = 15a^4 * cos(x^2-3x) * (2x-3)

dy/dx = 15(sin(x^2-3x))^4 * cos(x^2-3x) * (2x-3)

aw thanks ,, that's very nice of ye!!
yea this is the formula given in the book but my teacher does it in a totally different way ..:o so i am still trying to figure out what's the formula is about O_o ...

~Candy~

1huge1 wrote: »

Honours calculus I assume, Glad I dropped before we got to that.

In pass the chain rule never has anything like Sin in it.

yea, so you made the right decision ?
some of my mates dropped as well, because they don't want invest the time in maths :L
diff. was the 1st thing we did in 5th year ..
I am trying to do some revision now

JC 2K3 wrote: »

Well, there are 3 things going on here; a sine of an polynomial to the power of 5. So all you do is differentiate each of them and multiply them together (along with the constant(3)).

I've illustrated this with nice colours

3sin^5(x^2-3x) = 3*((sin(x^2-3x))^5)

3*(5(sin(x^2-3x)^4)*cos(x^2-3x)*(2x-3)

i remember you !!..thanks for helping me again
and thanks for the colours..I kinda get you ..the 'multiply them together (along with the constant(3))'
part totally makes sense to me ..^^
but May i ask

so
sin(x^2-3x) is one thing
and (x^2-3x) is another thing?
they share the(x^2-3x)??
diff. each you got the blue bit and the green bit..

or is it

sin(x^2-3x))
is one thing and if you diff it you got cos(x^2-3x)*(2x-3)

JC 2K3

~Candy~ wrote: »

so
sin(x^2-3x) is one thing
and (x^2-3x) is another thing?
they share the(x^2-3x)??
diff. each you got the blue bit and the green bit..

or is it

sin(x^2-3x))
is one thing and if you diff it you got cos(x^2-3x)*(2x-3)

Both are correct, just different ways of looking at it.

You can think of the expression as 3 different things if you like, i.e. ((sin(x^2-3x))^5) as one thing, (sin(x^2-3x)) as another thing, and (x^2-3x) as another thing, differentiate them all and multiply them all together as well as with the constant(3).

OR

You can think of ((sin(x^2-3x))^5) as one thing and (sin(x^2-3x)) as another thing, differentiate them both and multiply them all together as well as with the constant(3).
However, to differentiate (sin(x^2-3x)), you're going to have to think of (sin(x^2-3x)) as one thing and (x^2-3x) as another thing, differentiate them both and multiply them together.

So essentially you're doing exactly the same thing both times, just thinking about it slightly differently. In terms of how you learned the chain rule, the second way might be an easier way to think about it, but it's matter of preference really.

alan4cult

I very much overuse the chain rule.
Here is how I'd do this problem:

y=3([sin(x^2 - 3x)]^5)

LET u = sin(x^2 - 3x)

y = 3u^5
dy/du = 15u^4

u = sin(x^2 - 3x)

LET a = x^2 - 3x

u = sin a
du/da = cos a

a = x^2 - 3x
da/dx = 2x - 3

NOW MULTIPLY
dy/du * du/da * da/dx

(15u^4)(cos a)(2x - 3)

Now all back to same variable

15([sin(x^2 -3x)]^4)cos(x^2 - 3x)(2x - 3)

Overheal

1huge1 wrote: »

Honours calculus I assume, Glad I dropped before we got to that.

Im not. Did Ordinary but now in college we are exploring a lot of stuff you would otherwise see on the honors paper. And more.

Peleus

Ok, this is probably going to confuse things but i have a very easy mthod for those style q's. And i just got a new scanner so ill scan it, since typing it is such a bitch!

OK, my teacher teaches this: PTA.
this is used when you have a trig function to the power of something.

Mulitiply all these together:

1. Differentiate the power [bring the power forward and decrease the original power by 1]
2. Differentiate the Trig [forget about the power now and differentiate the trig (sin --> cos etc...)]
3. Differentiate the angle [diff inside the bracket.]

note: you have to include the angle part whereever you have a sin/cos in the answer.

works every time in one line.




If there is no power. Ie: Y= 3sin(x^2 - 3x)
then you just leave out the P part. So its just T.A

ie: Y= 3sin(x^2 - 3x)
dy/dx = 3 [ cos(x^2 - 3x)].[(2x-3)]

Hope this helps. PS: there are loads of different ways to do this question, i just find this easier and faster.

~Candy~

JC 2K3 wrote: »

Both are correct, just different ways of looking at it.

You can think of the expression as 3 different things if you like, i.e. ((sin(x^2-3x))^5) as one thing, (sin(x^2-3x)) as another thing, and (x^2-3x) as another thing, differentiate them all and multiply them all together as well as with the constant(3).

OR

You can think of ((sin(x^2-3x))^5) as one thing and (sin(x^2-3x)) as another thing, differentiate them both and multiply them all together as well as with the constant(3).
However, to differentiate (sin(x^2-3x)), you're going to have to think of (sin(x^2-3x)) as one thing and (x^2-3x) as another thing, differentiate them both and multiply them together.

So essentially you're doing exactly the same thing both times, just thinking about it slightly differently. In terms of how you learned the chain rule, the second way might be an easier way to think about it, but it's matter of preference really.

thanks very much..:pac:
well the last question before i leave this Q...
(i am such a 'annoying kid XD)hope you don't mind answer moi

the way i learned the chain rule, is bring down the power*the function^(power-1) * the differentiate function . do you think this would work for all cases? what if theres no power..

(sin(x^2-3x)) ..what rule did you use to diff this ?

alan4cult wrote: »

I very much overuse the chain rule.
Here is how I'd do this problem:

y=3([sin(x^2 - 3x)]^5)

LET u = sin(x^2 - 3x)

y = 3u^5
dy/du = 15u^4

u = sin(x^2 - 3x)

LET a = x^2 - 3x

u = sin a
du/da = cos a

a = x^2 - 3x
da/dx = 2x - 3

NOW MULTIPLY
dy/du * du/da * da/dx

(15u^4)(cos a)(2x - 3)

Now all back to same variable

15([sin(x^2 -3x)]^4)cos(x^2 - 3x)(2x - 3)

thanks!!! this looks handy.. maybe i should go and have a look of the formula that's given in the book...
:rolleyes:

JC 2K3

~Candy~ wrote: »

the way i learned the chain rule, is bring down the power*the function^(power-1) * the differentiate function . do you think this would work for all cases? what if theres no power..

If there's no power then why would you need to use the chain rule??

~Candy~ wrote: »

(sin(x^2-3x)) ..what rule did you use to diff this ?

The chain rule

(It's like the chain rule within the chain rule, if you get me)

~Candy~

Peleus wrote: »

Ok, this is probably going to confuse things but i have a very easy mthod for those style q's. And i just got a new scanner so ill scan it, since typing it is such a bitch!

OK, my teacher teaches this: PTA.
this is used when you have a trig function to the power of something.

Mulitiply all these together:

1. Differentiate the power [bring the power forward and decrease the original power by 1]
2. Differentiate the Trig [forget about the power now and differentiate the trig (sin --> cos etc...)]
3. Differentiate the angle [diff inside the bracket.]

note: you have to include the angle part whereever you have a sin/cos in the answer.

works every time in one line.

wow wow wow thank you! ^ ^
this rocks totally!!:D..
so in step 2 you just diff the sin and rewrite the angle..
yay...got it now toally, finally ...O_o

~Candy~

JC 2K3 wrote: »

If there's no power then why would you need to use the chain rule??


The chain rule

(It's like the chain rule within the chain rule, if you get me)

lol..thank you...again ( couldn't think of a better word to say)
:pya legend!! god i hope this Q come up in my LC paper ..

Davidius

Isn't that PTA method the way you're supposed to do them? Writing out the whole thing would be incredibly tedious. :pac:

MathsManiac

Davidius wrote: »

Isn't that PTA method the way you're supposed to do them? Writing out the whole thing would be incredibly tedious. :pac:

...but it only works if you have those particular types of functions nested in that particular way.

Peleus

Davidius wrote: »

Isn't that PTA method the way you're supposed to do them? Writing out the whole thing would be incredibly tedious. :pac:

yes but it only works if the question is something like this:

y= sinx
y= 5cos2x
y= 15cos(^2)5x
y= 3sin(^6)(5x+3)

etc... anyhting like that. if theres no power over the trig function, then just differentiate the trig function, multiply by differentiation of angle. ie: TA

ie: y=15cos(3x+2)

dy/dx= -15sin(3x+2).(3)

generally, you'd find yourself using it in the majority of trigonometry differentiation questions.

but if you got something like:

y= [sin(^3)3x].[cos3x]

then youd obviously use product rule or what ever methods there are for solving a product. I can't remember any of my diff cos i've just finished revising integration