Chain rule of multivariate calculus

[normal gulch]

WHY DO WE NEED TO USE CHAIN RULE WHEN WE CAN JUST SUBSTITUTE THE VALUES IN FIRST STEP?

[placid cipher]

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[hot thorn]

The chain rule implicitly does what you do

for instance, for x = r/s, since x depends on r, you have to differentiate r/s with respect to r

which is exactly what dw/dx * dx/dr is about

[normal gulch]

Oh

, rotate

[normal gulch]

This thing will gonna make disaster

[normal grotto]

A mathematician's nightmare

[normal gulch]

But why i am wrong here?

[normal grotto]

That's why you can't act like they're fractions

Because the $\frac{\partial w}{\partial x}$ disregards any changes in w wrt the other variables, thus multiplying them by $\frac{\partial x}{\partial r}$ via chain-rule doesn't take that in regard and thus does not equal $\frac{\partial w}{\partial r}$

[sly crescent]

Lumberdude #MakeWolfOwner

[normal grotto]

Hence $\frac{\partial w}{\partial x}\frac{\partial x}{\partial r}\neq \frac{\partial w}{\partial r}$

[normal gulch]

Is this also wrong?

[sly crescent]

Lumberdude #MakeWolfOwner

[normal grotto]

Hmmmm, no

the derivative of a variable wrt itself is just 1

[normal gulch]

But it's just small change in x /small change in x

∆x/∆x

[normal grotto]

Yeah and in the limit it is the derivative of x

it's not an actual fraction, it just comes down to being 1 like it were

Hope I didn't anger the actual mathmaticians with my explanation

[normal gulch]

But change is a number i think

Why it is not a fraction then

[normal grotto]

rate of change is a number if you're plugging in a value

Or, in this case if the derivative is constant, then it is a constant

Like, in the derivative of $f(x) = x^2$ it is not a number but a function. $f'(x)=2x$, which will have values at certain points

[sly crescent]

Lumberdude #MakeWolfOwner

[normal gulch]

dy/dx is a notation, not a number so we can't just do cancellation

Is this explanation correct?

[normal grotto]

Yes

I believe so

[normal gulch]

Ok

Thank you

[normal grotto]

No problem

[hot thorn]

You need to understand that in math, when we write $\frac{df}{dx}$, we do NOT treat $df$ and $dx$ like numbers

[sly crescent]

Rion

[hot thorn]

$\frac{df}{dx}$ is a function name with a very precise definition

[sly crescent]

Rion