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This question may look like a homework like question but I am only giving a example to express my problem :

Suppose we have an equation

$$ x^2=\cos\theta.$$

Taking the derivative to both sides I get

$$\frac{dx^2}{d\theta} = \sin\theta.$$

Now what $d(x^2)$ essentially means is that when I change $\theta$ to $\theta + d\theta$ what is the change in $x^2$. Now I know that

$$d(x^2)=2xdx$$

by as $\frac{dx^2}{dx}=2x$ and I can replace $d(x^2)$ from here. But my doubt Is how do we know that they both bring the same change $d(x^2)$ when $\theta$ changes to $\theta + d\theta$ and $x$ changes to $x +dx$. How do we know $d(x^2)$ is the same in both cases?

Sangchul Lee
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Naruto
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  • I don't understand your question, but maybe this helps: we have $$\dfrac {d x^2}{d \theta} = 2x \dfrac {d x}{d \theta}.$$ So we do take into account the change in $\theta$. – Ovi Jun 06 '20 at 04:10
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    You may think of this way: $df$ is an instrument that measures how the value of a function $f(x)$ changes as $x$ changes. So, even if the variable $x$ is restricted to a constraint, the relation $$df=f'(x),dx$$ continues to hold but the only difference is that now the variable $x$ can change only along that constraint. In your case, $x^2=\cos\theta$ induces $$2x,dx=-\sin\theta,d\theta,$$ and its interpretation is that two infinitesimal changes $2x,dx$ and $-\sin\theta,d\theta$ are coupled by the above equality. This ideal also aligns with the famous Chain Rule. – Sangchul Lee Jun 06 '20 at 04:18

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Good question. One way to think about differentiation which I think will clear some things up is to assume that there is a "standard candle" against which all differentiation is done.

That is, in your case you were noting that for the equation: $$ x^2 = \cos(\theta) $$ If you take the derivative with respect to $\theta$ you get $$ \frac{d(x^2)}{d\theta} = -\sin(\theta) $$ You noted that if you do the differentiation in the numerator you will get $d(x^2) = 2x\,dx$. But how do you know that these incremental changes "match up" so to speak?

The way to think about it is to assume that the differential operator is always in terms of some non-present standard candle. So, imagine a new variable, say $q$, which everything is ultimately dependent upon, but for which we do not have a formula. Therefore, we will say that $dq$ is some infinitely small difference (I usually use $\epsilon$ as the standard infinitesimal if I want to get concrete about it). You can think about the differential operator as always generating differentials which are relative to this standard candle. In fact, you can think about $dx$ and $d\theta$ as literally being relative values to our standard candle $dq$. Since $dq$ is a fixed value for our differential operator, then the relationships between $dx$ and $d\theta$ will hold true.

johnnyb
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