Undergrad Understanding the Chain Rule Equation: Explained with Examples

[AndrewGRQTF]

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?

 

[fresh_42]

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: $g(w)=2\cdot w = f(-w) = (-2)\cdot (-w)\,.$

 

[BvU]

$\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)$ - write them out using the definition of derivative

friend is right

 

[AndrewGRQTF]

$\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)$ - 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 $\Delta w \to 0$

 

[fresh_42]

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.

 

[AndrewGRQTF]

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?

 

[AndrewGRQTF]

$\frac {dg}{dw} \ne \frac {d}{dw} f(q,-w)$ - write them out using the definition of derivative

friend is right

How is my friend right?

 

[BvU]

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 !

 

[fresh_42]

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 $$\dfrac{dg}{dw}=\dfrac{df(h(w))}{dw}=-\dfrac{df(h(w))}{d(-w)}=-\dfrac{df(h)}{dh}=-\dfrac{df(w)}{dw}=\dfrac{df(-w)}{dw}$$

 

[AndrewGRQTF]

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

 

[Ray Vickson]

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 $h(w) = -w.$