Engineering Calculus Notes 258

# Engineering Calculus Notes 258 - 246 CHAPTER 3 REAL-VALUED...

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Unformatted text preview: 246 CHAPTER 3. REAL-VALUED FUNCTIONS: DIFFERENTIATION → Proposition 3.3.7 (Chain Rule R3 → R → R). Suppose f (− ) is a x → − , and g(y ) is a real-valued function of three variables, diﬀerentiable at v= x 0 → real-valued function of a real variable, diﬀerentiable at y = y0 = f (− 0 ). x → − ) is diﬀerentiable at − = − , and →→ Then the composition (g ◦ f )( x x x0 → − → −→ → ∇ (g ◦ f )(− 0 ) = g′ (y0 ) ∇ f (− 0 ) . x x Proof. This is formally very similar to the preceding proof. Let → → → △y = f (− 0 + △− ) − f (− 0 ) x x x then → −→ → → △y = ∇ f (− 0 ) · △− + δ △− x x x → − → where δ → 0 as △− → 0 . Note for future reference that x |△y | ≤ →→ −− ∇ f ( x 0) + δ → △− . x Now, → → g(f (− 0 + △− )) = g(y0 + △y ) x x = g(y0 ) + g′ (y0 ) △y + ε |△y | where ε → 0 as △y → 0. From this we can conclude → −→ → → → g(f (− 0 + △− )) − g(f (− 0 )) − g′ (y0 ) ∇ f (− 0 ) x x x x → = g′ (y ) δ △− + ε |△y | . x 0 → Taking absolute values and dividing by △− , we have x 1 →→ −− → − − → → − ′ → − g(f ( x 0 + △ x )) − g(f ( x 0 )) − g (y0 ) ∇ f ( x 0 ) △x |△y | ≤ g′ (y0 ) |δ| + ε − △→ x = g′ (y0 ) |δ| + ε →→ −− ∇ f ( x 0) + δ . Both terms consist of a bounded quantity times one that goes to zero as → − → △− → 0 , and we are done. x ...
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