A more rigorous argument


by MaxManus
Tags: argument, rigorous
MaxManus
MaxManus is offline
#1
Jan9-09, 03:02 AM
P: 297
Hey, in my Schaum' Outline Calculus it says Dx(ex) = ex
Let y = ex. Then ln(y) = x. By implicit differentiation, [tex]\frac{1}{y}[/tex]y' = 1
therefor y' = y = ex


For a more rigorous argument, let f(x) = ln(x) and f-1(y) = ey.
Note that f'(x) = [tex]\frac{1}{x}[/tex]. By Theorem 10.2(b).

(f-1)'(y) = [tex]\frac{1}{f'(f**-1(y))}[/tex],

That is Dy = [tex]\frac{1}{1/e**y}[/tex] = ey

10.2(b): Let f be one-to-one and continuous on the interval (a,b) Then:
If f'x(x0 is differentiable and f'(x0) != 0, then f-1 is
differentiable at y0 = f(x0) and (f-1)'(y0)
= [tex]\frac{1}{f'(x0}[/tex]

Now, my question is what does "more rigorous argument" mean and what makes the second one more rigorous.
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HallsofIvy
HallsofIvy is offline
#2
Jan9-09, 07:10 AM
Math
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Thanks
PF Gold
P: 38,882
The first argument uses the fact (your "10.2(b)") that if y= f(x) and x= f-1(y) then
[tex]\frac{dx}{dy}= \frac{1}{\frac{dy}{dx}}[/tex]
The second argument proves that is true for this particular function, as part of the proof.
MaxManus
MaxManus is offline
#3
Jan10-09, 10:41 AM
P: 297
Thanks


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