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Increment theorem

In nonstandard analysis, a field of mathematics, the increment theorem states the following: Suppose a function y = f(x) is differentiable at x and that Δx is infinitesimal. Then for some infinitesimal ε, where

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In nonstandard analysis, a field of mathematics, the increment theorem states the following: Suppose a function y = f(x) is differentiable at x and that Δx is infinitesimal. Then Δ y = f ( x ) Δ x + ε Δ x {\displaystyle \Delta y=f'(x)\,\Delta x+\varepsilon \,\Delta x} for some infinitesimal ε, where Δ y = f ( x + Δ x ) f ( x ) . {\displaystyle \Delta y=f(x+\Delta x)-f(x).}

If Δ x 0 {\textstyle \Delta x\neq 0} then we may write Δ y Δ x = f ( x ) + ε , {\displaystyle {\frac {\Delta y}{\Delta x}}=f'(x)+\varepsilon ,} which implies that Δ y Δ x f ( x ) {\textstyle {\frac {\Delta y}{\Delta x}}\approx f'(x)} , or in other words that Δ y Δ x {\textstyle {\frac {\Delta y}{\Delta x}}} is infinitely close to f ( x ) {\textstyle f'(x)} , or f ( x ) {\textstyle f'(x)} is the standard part of Δ y Δ x {\textstyle {\frac {\Delta y}{\Delta x}}} .

A similar theorem exists in standard Calculus. Again assume that y = f(x) is differentiable, but now let Δx be a nonzero standard real number. Then the same equation Δ y = f ( x ) Δ x + ε Δ x {\displaystyle \Delta y=f'(x)\,\Delta x+\varepsilon \,\Delta x} holds with the same definition of Δy, but instead of ε being infinitesimal, we have lim Δ x 0 ε = 0 {\displaystyle \lim _{\Delta x\to 0}\varepsilon =0} (treating x and f as given so that ε is a function of Δx alone).

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