# Having Trouble With A Simple Equation Involving Differentials

Given that $xyz = c$, show that $dz = -z\left(\frac{dx}{x}+\frac{dy}{y}\right)$. I'am not sure how to get started. How do i differentiate this equation and wrt what? I really would like to try this before posting it but unfortunately i'am unable to even get started.

Thank You

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Hint: $d(xyz)=yz\,dx+ xz\,dy+xy\,dz$ – Raymond Manzoni Aug 23 '12 at 9:58

Supposing $x,y,z>0$ and applying the $\log$ we have

$$\log(xyz)=\log c$$

$$\log x+\log y+\log z=\log c$$

$$\frac{1}{x}dx+\frac{1}{y}dy+\frac{1}{z}dz=0$$

$$dz=-z\left(\frac{dx}{x}+\frac{dx}{y}\right)$$

The obtained result is valid also if the above conditions on the signs are not valid.

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If you are working with real numbers, applying the log is a bad thing to do ; you are assuming that the constants and variables $x,y,z,c$ are positive, which I don't think would be clear with the context. Using the product rule for differentials is really the way to go. – Patrick Da Silva Aug 23 '12 at 10:01

The key to understanding differentials, at least in practical use, is as a most general form of differentiation -- one done with respect to no variables, or else to a parameter which we must assume every variable involved is a function of. This is the key to the statement $y=x \implies dy = dx$, and its (semantically, deeper) equivalent statement $\frac{dy}{dx} = 1$ .

So, $xyz = c \implies d(xyz) = d(c)$ , and we apply usual rules of differentiation, as though each variable invokes a chain rule: $$d(xyz) = xy~dz + x~dy~z + dx~yz ~~,~~d(c) = 0 \text{ as c is constant.}$$ So solve for $dz$, and you should arrive at the same solution.

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