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10.13 - Unfinished business from section 2.2 We saw last...

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Unfinished business from section 2.2: We saw last time that the continuous function { x sin 1/ x for x 0 and { { 0 for x = 0 is not differentiable at x =0. - 0.4 - 0.2 0.2 0.4 - 0.2 - 0.1 0.1 0.2 0.3 0.4 Now consider the function { x 2 sin 1/ x for x 0 and f ( x ) = { { 0 for x = 0.
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Is f differentiable at x =0? ..?. . How can we approach the problem? ..?. . Go back to the definition of the derivative. The derivative at x =0 is defined as lim h 0 [ f ( h )– f (0)]/ h = lim h 0 f ( h )/ h = lim h 0 h sin 1/ h = . .?. . 0, by the squeeze theorem of section 1.4: h sin 1/ h lies between | h | and –| h |, both of which go to 0 as h 0. Or argue it by going back to the ε , δ definition of limits: since | f ( h ) – f (0)| = | h sin 1/ h | = | h | |sin 1/ h | | h | = | h – 0|, for every > 0, there exists a > 0, namely = , such that every h satisfying 0 < | h – 0| < also satisfies | f ( h ) – f (0)| < .
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Section 2.4: The product and quotient rules The product rule: ( f g ) = f g + f g . That is: If f and g are differentiable at a , then fg is differentiable at a , and ( f g ) ( a ) = f ( a ) g ( a ) + f ( a ) g ( a ). Baby example:
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10.13 - Unfinished business from section 2.2 We saw last...

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