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I'm studying ordinary derivative equations and I'm trying to understand how partial derivatives work. I'm trying to guess it using those equations here:

$$z = ax + by$$ $$z'= \frac{\partial z}{\partial x} = a + by'$$

and: $$y = zx$$ $$y'=\frac{\partial y}{\partial x} = z'x + x$$

I've looked other posts that asked similar things and searched the internet for information but I can only find explanations of partial derivative of a function ($\frac{\partial f}{\partial x}$)

user2952272
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  • I'm not sure where the partial derivatives come in here. Aren't you just differentiating with respect to $x$ the funtions $z=z(x)$ and $y=y(x)$? The later with the product rule. – Karl Oct 03 '15 at 18:23
  • I think you might need to explain your notation. The prime tends to be used for $\mathrm d/\mathrm dx$, not $\mathrm\partial/\partial x$. This answer may or may not answer some of your questions. – joriki Oct 10 '15 at 20:51

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Partial derivatives work similar to the derivatives I assume you have worked before. For example, if you have a function $z = z(x_1,\dots,x_n)$ then, when you apply the partial derivative with respect to some coordinate, for example, $x_j$, what happens is that the other variables behave like constant's in one-variable-calculus so you just treat them as constants. For example:

$$y(z(x),x) = z(x)x$$

$$\frac{\partial y}{\partial x} = \frac{\mathrm{d} z}{\mathrm{d}x}(x)x + z(x)$$

Because in your examples seems like you have used $z = z(x)$. But have another example: $y(x,z,w) = x^2w + zw + \frac{1}{x}$ so if you derive with respect to x you will have

$$\frac{\partial y}{\partial x} = 2xw + 0 + -\frac{1}{x^2}$$

Can you understand what I did here? If you have trouble with the concept take any book for Calculus in multi-variable functions. I think that this was your question.

R.W
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