Chapter 9

Chapter 9 - 1.1 Theorem. Let m and n he integers; let n>...

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Unformatted text preview: 1.1 Theorem. Let m and n he integers; let n> 0. Let My) = (Efflm :93; 11> 0. Then h _'__s_ differentiable, and h'(y) = %(%)m.y"l ~ Proof. Steg _l_:_ We first prove the theorem in the case m = 1. Let f(x) = xn for X> 0. Then the inverse function to f, denoted g(y), is the nth root function. By the theorem on the derivative of an inverse function, g'(y) exists and I _. g (Y) " fn(x) I where x = g(y). Now f'(x) = nxn"l . Therefore 1 l g'(Y) = _ = —"—"——:— n xn 1 mgr-flu 1 1 91.x MW)“ (2/37)”l n y H 1 n --1 EW‘ Y - SteE g_._ We prove the theorem in general. If m = 0, it is trivial; Otherwise, we apply the chain rule. We have My) = WW“ ; then h' (Y) mew)“ [gray-11 53 <§mm- f1. 0 I.2 Once one has checked that the laws of exponents hold for rational exponents (Notes G), one can write this formula in a manner that is much easier to remember: Theorem. Let r 23 a rational constant; let h(x) = xr for x > 0. Then h, is differentiable and h'(x) = rxr‘l. we will give a different proof of this theorem later on, one which holds When r is an arbitrary real constant. ...
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This note was uploaded on 04/26/2009 for the course MATH CALC taught by Professor Brubaker during the Spring '09 term at MIT.

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Chapter 9 - 1.1 Theorem. Let m and n he integers; let n>...

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