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1st derivative

  • Thread starter merced
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  • #1
44
1
Differentiate the function: f(u) = e1/u
So, I used the chain rule and figured out that
f '(u) = (-u-2) e1/u

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

Why can't I pretend that 1/u is x and then say that
f '(x) = ex = e1/u

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

Answers and Replies

  • #2
2,063
2
merced said:
Differentiate the function: f(u) = e1/u
So, I used the chain rule and figured out that
f '(u) = (-u-2) e1/u

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

Why can't I pretend that 1/u is x and then say that
f '(x) = ex = e1/u

In other words, does the exponent always have to be "x" only, for f '(x) = ex 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 fucntion of a fucntion. Therefore [tex]\frac{df}{du} = \frac{df}{dx}\frac{dx}{du}[/tex]
 
  • #3
44
1
Oooh, ok, thanks
 

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