
#1
Jan909, 03:02 AM

P: 297

Hey, in my Schaum' Outline Calculus it says D_{x}(e^{x}) = e^{x}
Let y = e^{x}. Then ln(y) = x. By implicit differentiation, [tex]\frac{1}{y}[/tex]y' = 1 therefor y' = y = e^{x} For a more rigorous argument, let f(x) = ln(x) and f^{1}(y) = e^{y}. 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 D_{y} = [tex]\frac{1}{1/e**y}[/tex] = e^{y} 10.2(b): Let f be onetoone and continuous on the interval (a,b) Then: If f'x(x_{0} is differentiable and f'(x_{0}) != 0, then f^{1} is differentiable at y_{0} = f(x_{0}) and (f^{1})'(y_{0}) = [tex]\frac{1}{f'(x0}[/tex] Now, my question is what does "more rigorous argument" mean and what makes the second one more rigorous. 



#2
Jan909, 07:10 AM

Math
Emeritus
Sci Advisor
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. 



#3
Jan1009, 10:41 AM

P: 297

Thanks



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