Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
2  
It didn't disappear it is still in the $A$ at the right! –  Raymond Manzoni Oct 14 '12 at 21:44
1  
Do not loose track that it is an implicit derivative. The derived function actually appears in its derived expression. –  busman Oct 14 '12 at 21:47
    
I see now. Thanks everyone. –  John Roberts Oct 14 '12 at 21:51

3 Answers 3

up vote 0 down vote accepted

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.

share|improve this answer

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

share|improve this answer

It didn't disappear. Note that the $-jx\frac k z$ came from the chain rule, and the rest was left alone, hiding in the $A$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.