How do I differentiate e^-2x * y?

AI Thread Summary
To differentiate e^-2x * y, the product rule is applied, resulting in d/dx[e^-2x * y] = e^-2x * dy/dx + (-2e^-2x) * y. The notation dx does not imply that y becomes x; instead, dy/dx indicates the derivative of y with respect to x. There is a discussion about the merits of Leibniz notation versus Newton's notation, with some preferring Leibniz for its clarity in indicating the variable of differentiation. Overall, understanding the product rule and the notation is essential for correctly differentiating the expression.
James889
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Hi,

I need to differentiate the following \frac{d}{dx}[e^{-2x}y]

Does the `dx` mean that the y turns into x?

I know i have to use the product rule.
 
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Hi James889! :smile:

No, you just write "dy/dx" …

d/dx ( f(x)y) = f'(x)y + f(x)dy/dx :wink:
 
tiny-tim said:
Hi James889! :smile:

No, you just write "dy/dx" …

d/dx ( f(x)y) = f'(x)y + f(x)dy/dx :wink:

Hm?, what do you mean `just write`?

I can't think of anyone i hate more than Wilhelm Leibniz, for coming up with this stupid notation :mad:
 
James889 said:
Hm?, what do you mean `just write`?

I can't think of anyone i hate more than Wilhelm Leibniz, for coming up with this stupid notation :mad:

he he, get used to it! :biggrin:

(it's particularly useful when you start using substitution of variables in integals)

of course, you could write " y' " instead. :wink:
 
James889 said:
I can't think of anyone i hate more than Wilhelm Leibniz, for coming up with this stupid notation :mad:
Personally, I prefer Leibniz notation over Newton's notation as being more informative. The notation dy/dx is lots more informative than, say y' inasmuch as the Leibniz notation tells you exactly which variable with which differentiation is with respect to.
 

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