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Let's see I have the following equation

$$ x=1 $$

I take the derivate of both sides with respect to $x$:

$$ \frac{\partial }{\partial x} x = \frac{\partial }{\partial x}1 $$

Therefore, $1=0$. Clearly, that is not the right approach.

So what is the right way to think of $x=1$. What kind of object is it?

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If we're taking derivatives, that means the things on each side of the equals sign are functions. What you've demonstrated is the correct statement that if $f$ is the identity function $f(x) = x$ and $g$ is the constant function $g(x) = 1$, then $f\ne g$.

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You can take derivative of both sides of an identity not an equation.

For example $$\sin^2 x + \cos^2 x =1$$ is an identity, so we can differentiate to get $$2\sin x \cos x -2\sin x\cos x =0$$ or $$\cos 2x = \cos ^2 x - \sin ^2 x $$ gives $$-2\sin 2x = -2\sin x \cos x-2\sin x \cos x $$

Which is $$\sin 2x = 2\sin x \cos x$$

But you can not differentiate the equation $$\sin x =x$$ to get $$\cos x =1$$

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  • $\begingroup$ Inquiring: is it correct in saying that solving $\cos x=1$ lets you know when $y=\sin x$ and $y=x$ have the same slope, and this works for the general case? That is, solutions to $f'(x)=g'(x)$ tells you at which $x$ values $f$ and $g$ have the same slope? $\endgroup$ – Andrew Chin Nov 8 at 23:27
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Partial derivative requires a function as argument therefore if we assume, with little a abuse of notation, $x$ as a function $x= x(x,y,z,\ldots)$ and we states that $x$ is a constant function $x(x,y,z,\ldots)=1$ then

$$\frac{\partial }{\partial x} x(x,y,z,\ldots) = \frac{\partial }{\partial x}1=0$$

Note that of course it should be better to use a different notation for $x$ as a function to distinguish it form the variable $x$, that is for example $\bar x= \bar x(x,y,z,\ldots)$.

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Let me throw my hat as well; according to fundamental theorem of calculus, which I assume we all know, derivative is the result of integration,

$$\int_0^1\frac{d}{dx}xdx=\int_0^1dx=[x-x]_0^1=1-0=1$$

I prefer to think integration and derivatives as what they are, area approximations under the curve and slope. I think your approach does not consider that when x is a constant, the derivative is zero, so of course it does not make much sense.

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