- #1
MaxManus
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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 = exFor 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.
Let y = ex. Then ln(y) = x. By implicit differentiation, [tex]\frac{1}{y}[/tex]y' = 1
therefor y' = y = exFor 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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