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Mean-Value Theorem (Several Variables) 1 Mean-Value Theorem (Several Variables) THEOREM THE MEAN-VALUE THEOREM (SEVERAL VARIABLES) If f is differentiable at each point of the line segment ab , then there exists on that line segment a point c between a and b such that f (b) − f (a) = ∇f (c) · (b − a). Pr oof . As t ranges from 0 to 1, a + t(b − a) traces out the line segment ab. The idea of the proof is to apply the one-variable mean-value theorem to the function g(t) = f (a + t[b − a]), t ∈ [0, 1]. To show that g is differentiable on the open interval (0 , 1), we take t ∈ (0, 1) and form g(t + h) − g(t) = f (a + (t + h)[b − a]) − f (a + t[b − a]) = f (a + t[b − a] + h[b − a]) − f (a + t[b − a]) = ∇f (a + t[b − a]) · h[b − a] + o(h[b − a]). Since ∇f (a + t[b − a]) · h(b − a) = [ ∇f (a + t[b − a]) · (b − a)]h and the o (h(b − a)) term is obviously o (h), we can write g(t + h) − g(t) = [ ∇f (a + t[b − a]) · (b − a)]h + o(h). Dividing both sides by h , we see that g is differentiable and g (t) = ∇f (a + t[b − a]) · (b − a). The function g is clearly continuous at 0 and at 1. Applying the one-variable mean- value theorem to g, we can conclude that there exists a number t 0 between 0 and 1 such that g(1) − g(0) = g (t 0 )(1 − 0). Since g (1) = f (b), g(0) = f (a), and g (t 0 ) = ∇f (a + t 0 [b − a]) · (b − a), the above gives f (b) − f (a) = ∇f (a + t 0 [b − a]) · (b − a). March 17, 200 1

Chain Rule in multivariate

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