Chain rule equation help

  • #1
If we have an equation ##g (q,w) =f(q,-w)## and we want to find the derivative of that equation with respect to w, we would normally do $$\frac {dg}{dw} = \frac {d}{dw} f(q,-w) = \frac {df}{d(-w)} \frac {d(-w)}{dw} = -\frac {df}{d(-w)} $$ but my friend is saying that $$\frac {dg}{dw}= -\frac {df}{dw}$$ how is the last equation true?
 

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  • #2
fresh_42
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If we have an equation ##g (q,w) =f(q,-w)## and we want to find the derivative of that equation with respect to w, we would normally do $$\frac {dg}{dw} = \frac {d}{dw} f(q,-w) = \frac {df}{d(-w)} \frac {d(-w)}{dw} = -\frac {df}{d(-w)} $$ but my friend is saying that $$\frac {dg}{dw}= -\frac {df}{dw}$$ how is the last equation true?
Where has the minus in ##d(-w)## gone to? We have ## \frac {dg}{dw} =-\frac {df}{d(-w)}= \frac {df}{dw}\,.##

Make a simple example: ##g(w)=2\cdot w = f(-w) = (-2)\cdot (-w)\,.##
 
  • #3
BvU
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##\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)## - write them out using the definition of derivative

friend is right
 
  • #4
##\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)## - write them out using the definition of derivative

friend is right

But how is ##\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)##? It is given that ##g(q,w) = f(q,-w)## so we just take the derivative of both sides?

Writing out the derivative we have $$\frac{g(q,w+\Delta w) - g(q,w)}{\Delta w} = \frac{f(q,-(w+\Delta w)) - f(q,w)}{\Delta w}$$ as ##\Delta w \to 0##
 
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  • #5
fresh_42
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But how is ##\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)##?
It is not unequal! If ##f=g## then ##\dfrac{df}{dw}=\dfrac{dg}{dw}\,.##

First we write ##f(q,-w)=f(q,h(w))## with ##h(w)=-w\,##. Then calculate again ##\dfrac{df}{dw}\,##.
Your friend set ##dw=d(h(w))=d(-w)## which is obviously wrong.

The confusion is due to bad notation, i.e. dropping the variables of the functions. Write it with ##h(.)## and do not drop them anywhere.
 
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  • #6
It is not unequal! If ##f=g## then ##\dfrac{df}{dw}=\dfrac{dg}{dw}\,.##

First we write ##f(q,-w)=f(q,h(w))## with ##h(w)=-w\,##. Then calculate again ##\dfrac{df}{dw}\,##.
Your friend set ##dw=d(h(w))=d(-w)## which is obviously wrong.

The confusion is due to bad notation, i.e. dropping the variables of the functions. Write it with ##h(.)## and do not drop them anywhere.

Ok, thanks for your replies. Why is BvU saying what he's saying?
 
  • #7
##\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)## - write them out using the definition of derivative

friend is right

How is my friend right?
 
  • #8
BvU
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Make a simple example: ##g(w)=2\cdot w = f(-w) = (-2)\cdot (-w)\,.##

I'm having a bad night. My simple example:
$$g(w) = w = f(-w)$$ so ##g(1) = 1## and ## g(-1) =-1## and I was looking at ##f'## and ##g'## -- WRONG ! o:) o:)
 
  • #9
fresh_42
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After I do that and get $$\frac{df}{dw} = \frac{df}{dh} \frac{dh}{dw} = -\frac{df}{dh}$$ what do I do, why did you rewrite it that way?
I rewrote it, because the minus sign at ##w## is the cause of evil here. By writing it as what it is, namely a function itself whose derivative affects the result, I minimize the chance of mumbo-jumbo and brooming it somewhere under the carpet. The situation is (forget the ##q##, esp. as nobody used partial derivatives here!):
##g(w) = f(-w)=f(h(w))## and ##d(-w) = d(h(w)) = -dw## hence $$\dfrac{dg}{dw}=\dfrac{df(h(w))}{dw}=-\dfrac{df(h(w))}{d(-w)}=-\dfrac{df(h)}{dh}=-\dfrac{df(w)}{dw}=\dfrac{df(-w)}{dw}$$
 
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  • #10
I rewrote it, because the minus sign at ##w## is the cause of evil here. By writing it as what it is, namely a function itself whose derivative affects the result, I minimize the chance of mumbo-jumbo and brooming it somewhere under the carpet. The situation is (forget the ##q##, esp. as nobody used partial derivatives here!):
##g(w) = f(-w)=f(h(w))## and ##d(-w) = d(h(w)) = -dw## hence $$\dfrac{dg}{dw}=\dfrac{df(h(w))}{dw}=-\dfrac{df(h(w))}{d(-w)}=-\dfrac{df(h)}{dh}=-\dfrac{df(w)}{dw}=\dfrac{df(-w)}{dw}$$

Thanks a lot. I learned something important: not to forget about the argument of a function
 
  • #11
Ray Vickson
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If we have an equation ##g (q,w) =f(q,-w)## and we want to find the derivative of that equation with respect to w, we would normally do $$\frac {dg}{dw} = \frac {d}{dw} f(q,-w) = \frac {df}{d(-w)} \frac {d(-w)}{dw} = -\frac {df}{d(-w)} $$ but my friend is saying that $$\frac {dg}{dw}= -\frac {df}{dw}$$ how is the last equation true?
$$\frac{d}{dw} f(q,-w) = \frac{d}{d(-w)} f(q,-w) \cdot \frac{d(-w)}{dw} = - f'(q,-w),$$
where ##f'(q,x) = (d/dx) f(q,x),## so ##f'(q,-w) = \left. f'(q,x) \right|_{x=-w}##

Basically, you just need to apply the chain rule to ##f(q,h(w))##, with ##h(w) = -w.##
 

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