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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 mgrﬂu 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)“ [gray11 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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 Spring '09
 BRUBAKER
 Calculus, Derivative, Exponentiation, Inverse function, Function composition

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