Why Chain Rule for Differentiating f(u) = e1/u?

[merced]

Differentiate the function: f(u) = e^1/u
So, I used the chain rule and figured out that
f '(u) = (-u^-2) e^1/u

My question is, why do you have to use the chain rule?
I know that if f(x) = e^x
then f '(x) = e^x

Why can't I pretend that 1/u is x and then say that
f '(x) = e^x = e^1/u

In other words, does the exponent always have to be "x" only, for f '(x) = e^x to work?

 

[neutrino]

Differentiate the function: f(u) = e^1/u
So, I used the chain rule and figured out that
f '(u) = (-u^-2) e^1/u

My question is, why do you have to use the chain rule?
I know that if f(x) = e^x
then f '(x) = e^x

Why can't I pretend that 1/u is x and then say that
f '(x) = e^x = e^1/u

In other words, does the exponent always have to be "x" only, for f '(x) = e^x to work?

The derivative e^u with respect to u is e^u and the derivative e^x with respect to x is e^x, and it does not matter what alpahbet you choose to denote the variable with. It's a dummy.

But in the problem you have posted, if you assume that x = 1/u, then the function is f(x(u)) [since x is now a function of u], and that is why you use the chain rule. You assume it to be a function of a function. Therefore \frac{df}{du} = \frac{df}{dx}\frac{dx}{du}

 

[merced]

Oooh, ok, thanks