Proofs_MATH1500.pdf

# Proofs_MATH1500.pdf - PROOFS MATH 1500 PROOFS Liliana...

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PROOFS MATH 1500 - PROOFS Liliana Menjivar Liliana Menjivar MATH 1500 - PROOFS

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PROOFS Table of contents PROOFS 1. Differentiable functions are continuous. 2. Constant Multiple Rule. 3. Sum Rule. 4. Product Rule. 5. Derivative of sin( x ) is cos( x ). 6. If f 0 ( x ) = 0, then f is constant. 7. If f 0 ( x ) > 0 , then f is increasing. 8. If f 0 ( x ) < 0 , then f is decreasing. Liliana Menjivar MATH 1500 - PROOFS
PROOFS 1. Differentiable functions are continuous. Theorem If f is differentiable at x = a , then f is continuous at x = a . Liliana Menjivar MATH 1500 - PROOFS

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PROOFS 1. Differentiable functions are continuous. Proof. We need to prove that lim x a f ( x ) = f ( a ) . By assumption, f is differentiable at a , so f 0 ( a ) = lim x a f ( x ) - f ( a ) x - a exists. Then lim x a f ( x ) = lim x a f ( x ) - f ( a ) + f ( a ) = lim x a ( f ( x ) - f ( a )) + lim x a f ( a ) = lim x a f ( x ) - f ( a ) · x - a x - a + f ( a ) = lim x a f ( x ) - f ( a ) x - a · lim x a ( x - a ) + f ( a ) = f 0 ( a ) · 0 + f ( a ) = f ( a ) . Therefore, f is continuous at a . Liliana Menjivar MATH 1500 - PROOFS
PROOFS 2. Constant Multiple Rule. Theorem (Constant Multiple Rule.) If c is a constant, and f is a differentiable function, then d dx c f ( x ) = c d dx f ( x ) . Alternate notation. ( c f ) 0 = c f 0 . Liliana Menjivar MATH 1500 - PROOFS

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PROOFS 2. Constant Multiple Rule. Proof. Let G ( x ) = cf ( x ). Then G 0 ( x ) = ( cf ( x )) 0 = lim h 0 G ( x + h ) - G ( x ) h = lim h 0 c f ( x + h ) - c f ( x ) h = lim h 0 c [ f ( x + h ) - f ( x )] h = c lim h 0 f ( x + h ) - f ( x ) h = c f 0 ( x ) . Liliana Menjivar MATH 1500 - PROOFS
PROOFS 3. Sum Rule.

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