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If $z = x + D(x + y)$ and I let $g(x,y) = x + y$

Would I be right in saying that :

$z = x + f(g(x))$

and

$\frac{dz}{dx} = 1 + \frac{df}{dg}\cdot\frac{dg}{dx}$?

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D(x + y) is a function of one variable. g(x,y) is a substitution to apply chain rule –  jedd Jan 25 '11 at 18:36
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2 Answers

up vote 1 down vote accepted

I assume $D(.)$ is a function not a constant.

Then yes: $$ \frac{\partial }{\partial x}z(x,y) = 1 + \frac{\partial }{\partial g}D(g(x,y)) \frac{\partial }{\partial x}g(x,y)$$

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I think you meant $z=x+D(g(x,y))$. If $D$ is a function of one variable, then you can talk about the partial with respect to $x$; you would be correct in saying that $$\frac{\partial z}{\partial x} = 1 + \frac{dD}{dg}\frac{\partial g}{\partial x}.$$ That is, $$\frac{\partial z}{\partial x} = 1 + D'(x+y)$$ because $\frac{\partial g}{\partial x} = \frac{\partial}{\partial x}(x+y) = 1$.

Notice that for functions of one variable, we use $d$ and $dx$, but for functions of several variables you are talking about the partial derivative, $\partial$ and $\partial x$.

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Thanks for quick and clear response –  jedd Jan 25 '11 at 18:46
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