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$$ A=(A_0/z)\exp[-jk(x^{2}+y^{2})/2z] $$ $$ \frac{\partial{A}}{\partial{x}} = -jxA\frac{k}{z} $$ Can anybody explain why this is the case? I thought that exponential functions never disappeared when one does derivatives.

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    $\begingroup$ It didn't disappear it is still in the $A$ at the right! $\endgroup$ Commented Oct 14, 2012 at 21:44
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    $\begingroup$ Do not loose track that it is an implicit derivative. The derived function actually appears in its derived expression. $\endgroup$
    – busman
    Commented Oct 14, 2012 at 21:47
  • $\begingroup$ I see now. Thanks everyone. $\endgroup$ Commented Oct 14, 2012 at 21:51

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Well, it doesn't disappear. What you didn't write is, that A is a function depending on $x$. Actually you have

$\partial_x A = -jk\frac{k}{z}A(x)$

and the $\exp$ is included in your function A(x). This is the prototype of an ordinary differential equation.

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Let $B=e^x$. Then $B'=B$. Has the exponential dísappeared?

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