Derivative using the chain rule

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  • Thread starter tomwilliam
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  • #1
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I'm coming back to maths (calculus of variations) after a long hiatus, and am a little rusty. I can't remember how to do the following derivative:

##
\frac{d}{d\epsilon}\left(\sqrt{1 + (y' + \epsilon g')^2}\right)
##
where ##y, g## are functions of ##x##

I know I should substitute say ##u = 1 + (y' + \epsilon g')^2##
then use the chain rule, ## \frac{\partial\sqrt{u}}{\partial x} \frac{\partial x}{\partial \epsilon}##

But now I'm a little stuck. Can anyone help with a pointer?
I know what the final answer is, but can't get there.
Thanks
 

Answers and Replies

  • #2
PeroK
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I'm coming back to maths (calculus of variations) after a long hiatus, and am a little rusty. I can't remember how to do the following derivative:

##
\frac{d}{d\epsilon}\left(\sqrt{1 + (y' + \epsilon g')^2}\right)
##
where ##y, g## are functions of ##x##

I know I should substitute say ##u = 1 + (y' + \epsilon g')^2##
then use the chain rule, ## \frac{\partial\sqrt{u}}{\partial x} \frac{\partial x}{\partial \epsilon}##

But now I'm a little stuck. Can anyone help with a pointer?
I know what the final answer is, but can't get there.
Thanks
You have a function ##f(u(\epsilon))## and the chain rule is:
$$\frac{df}{d\epsilon} = \frac{df}{du}\frac{du}{d\epsilon}$$ Note that ##x## and##y## don't come into this.
 
  • #3
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Ok, thanks, I see where my mistake came in.
So now I have
##
\frac{d\sqrt{u}}{du}=(1/2)u^{-1/2}
##
and
##
\frac{du}{d\epsilon}= 2(y' + \epsilon g')
##

so ## \frac {df}{d\epsilon} = (1/2)u^{-1/2}\times 2(y' + \epsilon g') ##

but I seem to be a factor of ##g'## out.
 
  • #4
PeroK
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##
\frac{du}{d\epsilon}= 2(y' + \epsilon g')
##
You need to take more care over that step.
 
  • #5
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I can't remember how to do the following derivative:

##
\frac{d}{d\epsilon}\left(\sqrt{1 + (y' + \epsilon g')^2}\right)
##
where ##y, g## are functions of ##x##

I know I should substitute say ##u = 1 + (y' + \epsilon g')^2##
then use the chain rule, ## \frac{\partial\sqrt{u}}{\partial x} \frac{\partial x}{\partial \epsilon}##
Since it's not given that ##\epsilon## is a function of x, there is no need for partial derivatives here. As @PeroK already noted, y and g (and therefore y' and g') don't enter into the calculation at all. They can all be considered to be constants, as far as the differentiation goes.
With your substitution, ##u = 1 + (y' + \epsilon g')^2 = h(u)##,
the chain rule would look like this: ##\frac d {du}(u^{1/2}) \cdot \frac {du}{d\epsilon}##.
 
  • #6
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Thanks to both of you...I understand it now.
 

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