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I know these questions may seem trivial, but I can't understand the following thing:

Say there's a function $F(x,y,y')=x+y+y', where ~y(x)=x^3$.

$\frac{\partial F}{\partial x} = 1+ \frac{\partial (y)}{\partial x}+\frac{\partial (y')}{\partial x}$

here, on the textbook, it states that $\frac{\partial (y)}{\partial x}=0 ~and ~\frac{\partial (y')}{\partial x}=0$ ...

At first glance, this seemed obvious because the terms y and y' don't have 'x' in them. However, since $y=x^3$, shouldn't $\frac{\partial (y)}{\partial x}=\frac{\partial(x^3)}{\partial x} =3x^2 $ and so on?

Also, in the case above, what is $\frac{\partial F}{\partial y}$?

Shouldn't the $\partial y'/\partial y$ not equal zero, as y' is related to y?

Danny Han
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1 Answers1

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Roughly speaking, taking a partial derivative means that we treat all variables of the function as independent variable and we are taking the derivative with respect to one of those independent variables. Say, for a function $F(x,y,y')$ in general or for your example $F(x,y,y')=x+y+y'$ specifically, we treat it as a function of the three independent variables $x$, $y$, and $y'$, and therefore there are three separate partial derivatives: $$\frac{\partial F}{\partial x}=1, \quad \frac{\partial F}{\partial y}=1, \quad\frac{\partial F}{\partial y'}=1.$$ And when treating all three of them as independent variables, then of course, the derivative of any one with respect to another is zero.

However, if one of these variables in turn depends on some other variables, then the multivariate Chain Rule kicks in. What you calculated in your question, where $y=x^3$, is not a partial derivative at all, but the total derivative with respect to $x$. It would even be denoted differently: $$\frac{\mathrm{d}F}{\mathrm{d}x}=\frac{\partial F}{\partial x}\cdot\frac{\mathrm{d}x}{\mathrm{d}x}+\frac{\partial F}{\partial y}\cdot\frac{\mathrm{d}y}{\mathrm{d}x}+\frac{\partial F}{\partial y'}\cdot\frac{\mathrm{d}y'}{\mathrm{d}x}=1\cdot1+1\cdot3x^2+1\cdot6x=3x^2+6x+1.$$

zipirovich
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