MATH 180 Study Guide - Final Guide: Maxima And Minima

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13 Dec 2018
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Problem 1 solution: find derivatives of the following functions: (a) f (x) = tan 1( x2 + 1) (b) f (x) = ln(cid:18) 2x x2 + 1(cid:19) (c) f (x) = xsin(x) Solution: (a) the derivative is computed using the chain rule twice. f (x) = 2 x2 + 1 dx (x2 + 1) (b) we begin by rewriting the function using rules of logarithms. ln(cid:18) 2x x2 + 1(cid:19) = ln(2x) ln(x2 + 1) = x ln(2) ln(x2 + 1) Thus, the derivative is f (x) = d dx x ln(2) . 1 d dx ln(x2 + 1) d (x2 + 1) = ln(2) x2 + 1 dx x2 + 1 2x. 1 (c) to di erentiate the function we rewrite it as xsin(x) = eln xsin(x) The derivative is then f (x) = d dx esin(x) ln(x) Problem 2 solution: use implicit di erentiation to nd dy dx where tan(x2y) = 4y.

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