# tut1 - 1 Tutorial 1 Comments Q1 f(x = x|x| To nd f(0 we...

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Tutorial 1 Comments Q1 f ( x ) = x | x | . To nd f 0 (0) , we have by de nition f 0 (0) = lim x 0 f ( x ) - f (0) x = lim x 0 x | x | if the limit EXISTS. Now consider the function g ( x ) = x | x | . This function is continuous for all x R . Recall that for a function g ( x ) which is continuous for all x R , by de nition, lim x c g ( x ) = g ( c ) , for all c R . Therefore, f 0 (0) = lim x 0 x | x | = lim x 0 g ( x ) = g (0) = 0 . Recall too that the reason for which the function h ( x ) = | x | is not di erentiable at x = 0 is that f 0 (0) = lim x 0 f ( x ) - f (0) x = lim x 0 | x | x DOES NOT EXIST ! We can see it in two ways : 1) The function | x | x is NOT continuous. (There is a jump at x=0 since | x | x = - 1 for x < 0 and | x | x = 1 for x > 0 . 2) As stated in the lecture notes, the left-hand and right hand limits are not equal, and thus the limit does not exist. (we can see this as also an intepretation of the fact that | x | x is not continuous) Q2 When encountered with problems such as nding the derivative of a product of functions (more than 3 say), f ( x ) = g ( x ) h ( x ) j ( x ) k ( x ) , an alternative way to approach the problem is to take ln on both sides of the equation so that the product becomes a sum. ln

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