$ 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.
Help With Partial Derivative
0
$\begingroup$
calculus
derivatives
-
0I see now. Thanks everyone. – 2012-10-14
2 Answers
0
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.
0
Let $B=e^x$. Then $B'=B$. Has the exponential dÃsappeared?