2007-10-05 Proof of the Chain Rule

# 2007-10-05 Proof of the Chain Rule - g(xo 611 = g(Xo)6X...

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y Proof of the Chain Rule oj Our goal is to show that if I (u) is a differentiable function of II and u = g (x) is a differentiablefunctionof x, then the composite y = I(g(x» is a differentiable function of x. More precisely, if g is differentiable at Xo and I is differentiable at g(xo), then the composite is differentiable. at Xo and dY / I ' , - = (g(xo))' g (xo). dx x=x, Let .6.x be an increment in x and let .6.u and .6.y be the corresponding increments in u and y. dY ! ..6.y - = hm -, dx x=x, ~x~O.6.x so our goal is to show that this limit is f'(g(xo»
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Unformatted text preview: . g'(xo). 611 = g'(Xo)6X + £,6X = (g'(xo) + CI)~X, where c, -+ 0 as 6X-+ O. Similarly, 6y = f'(Llo)6U + £2 6U = (j'(uo) + E:)?U, where C2-+ 0 as 6u-+ O. Notice also that ~U __ 0 as _'.x-O. Combii1ir:g ,he equations for 611 and ~y gives so Since c, and c2 go to zero as ~x goes to zero, three of the four t:?rms on the rifh[ vanish in the limit, ]eaving . 6y I ' ' ( ' I .. , hm -= (LlO)g xo) = (5(x(,)). g (xc). ~x-O 6X...
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