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[SOLVED] chain rule problem and choice of notation

DeusAbscondus

Active member
Jun 30, 2012
176
I have attached a pdf setting forth my question.
This is a write up of a lesson i just had on yourtutor, in which i think the tutor might have made an error: this is a direct quote from the whiteboard:
$Let g(x)=2x, f(y)=e^y\Rightarrow(fog)(x)=f(g(x))=f(2x)=e^{2x}$$\\Now $$(fog)'(x)=f'(g(x))=f'(g(x)).g'(x)$

$f'(y)=e^{y}$
$g'(x)=2$
$(fog)'(x)=f'(2x).2=2e^{2x}$

My approach to solving problems like this has been unsystematic; he has tried to get me to think about these chain-rule problems in terms of function notation, rather than the unseemly parade of variables I trade in, as can be seen below

$\text{Now if: }y=x^2.e^{-x} \text{then let: }f(x)=x^2; g(x)=e^{-x};y'=f'(x).g'(x)=2x.-e^{-x}$

Would someone kindly comment critically and with explanations on the two formats/usages and their relative strengths/weaknesses as a modus operandi for all such problems, at the level of beginner.
Thanks,
DeusAbs
 
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CaptainBlack

Well-known member
Jan 26, 2012
890
I have attached a pdf setting forth my question.
This is a write up of a lesson i just had on yourtutor, in which i think the tutor might have made an error: this is a direct quote from the whiteboard:
$Let g(x)=2x, f(y)=e^y\Rightarrow(fog)(x)=f(g(x))=f(2x)=e^{2x}$$\\Now $$(fog)'(x)=f'(g(x))=f'(g(x)).g'(x)$
That last line cannot be right as it stands, or rather the middle term is wrong.

What may be meant is:

$$(fog)'(x)=(f(g(x)))'=f'(g(x)).g'(x)$$

but that is not right either as the prime acts on a function not a function value.

CB
 

HallsofIvy

Well-known member
MHB Math Helper
Jan 29, 2012
1,151
I have attached a pdf setting forth my question.
This is a write up of a lesson i just had on yourtutor, in which i think the tutor might have made an error: this is a direct quote from the whiteboard:
$Let g(x)=2x, f(y)=e^y\Rightarrow(fog)(x)=f(g(x))=f(2x)=e^{2x}$$\\Now $$(fog)'(x)=f'(g(x))=f'(g(x)).g'(x)$

$f'(y)=e^{y}$
$g'(x)=2$
$(fog)'(x)=f'(2x).2=2e^{2x}$

My approach to solving problems like this has been unsystematic; he has tried to get me to think about these chain-rule problems in terms of function notation, rather than the unseemly parade of variables I trade in, as can be seen below

$\text{Now if: }y=x^2.e^{-x} \text{then let: }f(x)=x^2; g(x)=e^{-x};y'=f'(x).g'(x)=2x.-e^{-x}$
Would someone kindly comment critically and with explanations on the two formats/usages and their relative strengths/weaknesses as a modus operandi for all such problems, at the level of beginner.
Thanks,
DeusAbs
All I can say is that your "method" is completely wrong and gives a wrong result. What you have written as your function, [tex]y= x^2e^{-x}[/tex] is a product, not a composition. And the product rule is (fg)'= f'g+ fg', NOT f'g'. The derivative of [tex]y= x^2e^{-x}[/tex] is [tex]y'= (x^2)'e^{-x}+ x^2(e^{-x})'= 2xe^{-x}+ x^2e^{-x}= e^{-x}(x^2+ 2x)[/tex].

If you meant [tex]y= e^{-x^2}[/tex] then it would be [tex]y'= e^{-x^2}(-x^2)'= e^{-x^2}(-2x)= -2xe^{-x^2}[/tex]
 
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DeusAbscondus

Active member
Jun 30, 2012
176
All I can say is that your "method" is completely wrong and gives a wrong result. What you have written as your function, [itex]y= x^2e^{-x}[/itex] is a product, not a composition. And the product rule is (fg)'= f'g+ fg', NOT f'g'. The derivative of [itex]y= x^2e^{-x}[/itex] is [itex]y'= (x^2)'e^{-x}+ x^2(e^{-x})'= 2xe^{-x}+ x^2e^{-x}= e^{-x}(x^2+ 2x)[/itex].

If you meant [itex]y= e^{-x^2}[/itex] then it would be [itex]y'= e^{-x^2}(-x^2)'= e^{-x^2}(-2x)= -2xe^{-x^2}[/itex]
Thanx for the bracing cold slap of reality: I learned from it!
D'abs.
 
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DeusAbscondus

Active member
Jun 30, 2012
176
That last line cannot be right as it stands, or rather the middle term is wrong.

What may be meant is:

$$(fog)'(x)=(f(g(x)))'=f'(g(x)).g'(x)$$

but that is not right either as the prime acts on a function not a function value.

CB
Thanks, I knew it couldn't be rigth but couldn't see why.
D'abs
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
If you are going to be unfriendly and as unfinessed as a blunt battle-axe, at least be clear!
Tsk, tsk: look at your mess~!

D'abs
Don't take it personally. HallsofIvy has been helping on lots of sites for many, many years and has probably come across thousands upon thousands of lazy, ungrateful students in that time. When you help out on these kind of sites for a while you start to shorten your answers and just get straight to the point. I'm almost positive he didn't mean to insult you and is really trying to help you. We've given him his badges for a reason :)
 

DeusAbscondus

Active member
Jun 30, 2012
176
Don't take it personally. HallsofIvy has been helping on lots of sites for many, many years and has probably come across thousands upon thousands of lazy, ungrateful students in that time. When you help out on these kind of sites for a while you start to shorten your answers and just get straight to the point. I'm almost positive he didn't mean to insult you and is really trying to help you. We've given him his badges for a reason :)
Big pause for thought.....
You are right: I *am* a tetchy creature: the question could have been a lot more carefully prepared; in that way I might have had no need to post it!!!

Suitably chastened, informed, another bit of the wild-man tamed and fitter for civilization.

thanks!
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,052
Big pause for thought.....
You are right: I *am* a tetchy creature: the question could have been a lot more carefully prepared; in that way I might have had no need to post it!!!

Suitably chastened, informed, another bit of the wild-man tamed and fitter for civilization.

thanks!
What I meant most of all is it's understandable how you felt and reacted. I visit other non-math forums and see how rude and inconsiderate people are to each other on the internet. If someone at MHB is legitimately being rude it will be dealt with and you always have the right to bring a post to our attention. I'll close this thread in a day or so if there are no more math related comments, since I'm kind of derailing it into a feedback thread but just wanted to make sure saw that I wasn't trying to reprimand you at all.
 

DeusAbscondus

Active member
Jun 30, 2012
176
What I meant most of all is it's understandable how you felt and reacted. I visit other non-math forums and see how rude and inconsiderate people are to each other on the internet. If someone at MHB is legitimately being rude it will be dealt with and you always have the right to bring a post to our attention. I'll close this thread in a day or so if there are no more math related comments, since I'm kind of derailing it into a feedback thread but just wanted to make sure saw that I wasn't trying to reprimand you at all.
Absolutely.
I took your comments in exactly this sense Jameson. You handled the matter with tact and sensitivity to all, which is what transformed a possibly acrimonious moment for me and others into a deepened understanding of:
-why I am here;
-the specialness of the place; and
-the need to sometimes wait 6 hours before responding :)

Thanks for the extra clarification though.
Deus Abs