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18.01 SPRING 2008 PROBLEM SET 2 SOLUTIONS Graded problems, Part A See attached photocopies. Graded problems, Part B 1. (a) Let y = xp/q , so by definition y q = xp . Differentiate both sides of this with respect to x, obtaining qy q−1 dy = pxp−1 . dx Note that we’re allowed to use the power rule for this because both p and q are assumed to be dy , this becomes positive integers. Solving for dx p dy p xp−1 p p p xp−1 p xp−1 = = xp−1−(p− q ) = xp/q−1 . = = p q−1 p/q q−1 (q−1) dx qy q (x ) q xq q q (b) Assume n is a positive rational number and y = x−n . Then yxn = 1, and differentiating both sides with respect to x yields dy n x + ynxn−1 = 0. dx Here we’ve used the power rule to differentiate xn since n is positive. Solving for dy dx , we find yxn−1 x−n xn−1 dy = −n n = −n = −nx−n−1 . dx x xn The upshot is that from now on you can feel free to use the power rule without worrying whether n is positive, negative, rational or whatever! d 2. The 4th derivative of sin x is itself, thus the same is true for dx n sin x whenever n is any multiple of 4. For sin ax we have to use the chain rule, which introduces a new factor of a in front every time d2 dn 2 n we differentiate, i.e. dx 2 sin ax = −a sin ax and so forth; in particular, dxn sin ax = a sin ax if n is divisible by 4. Since 172 is divisible by 4, we have n d2 d172 d2 172 d174 sin ax = sin ax = a sin ax = −a174 sin ax. dx174 dx2 dx172 dx2 1