# X 2 3 5 y f x x g x y 3 x 2 2 x y f x g x y 3 x 2 1 6

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x 2 + 3 5. y = f ( x ) x g ( x ) y = (3 x 2 )(2+ x ) y’ = f ( x ) g ’( x ) + y’ = (3 x 2 )(1) + = 6 x 2 + 3 x 3 g ( x ) f ’( x ) y’ (2+ x )(6 x ) =12 x + 9 x 2 6. y = g f ( ( x x ) ) y = (1 ( x + + 2 4 x ) ) y’ = [ g ( x ) f ’( x ) – y’ = ( x + 7 4) 2 f ( x ) g ’( x )]/[ g ( x )] 2 7. y = f ( u ( x )) y = (3 x + 4) 2 , with y’ = d d u y d d u x y’ = 2(3 x + 4)(3) u = 3 x +4, = 18 x + 24 y = u 2 8. y = ln x y = ln x y’ = 1/ x y’ = 1/ x 9. y = e x y = e x y’ = e x y’ = e x TABLE M.10-1 Basic Formulas for Derivatives

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firms want to maximize their profits. Similarly, a firm may want to find the level of out- put that minimizes its cost per unit of producing a good. When we can represent a vari- able like profit, say, as a continuously differentiable (or “smooth”) function of another variable (such as a firm’s output), we can use calculus to find the level (or levels) of out- put at which profit is maximized. Figure M.10-2 lets us see the conditions under which a maximum or a minimum will occur. Figure M.10-2(a) has the graph of the function y = x 2 – 4 x + 10, and Figure M.10- 2(b) has the graph of the function y = 2 + 4 x x 2 . Figure M.10-2(a) reaches a minimum at the point (2, 6), while Figure M.10-2(b) reaches a maximum at (2,6). In order to understand why, we need to calculate the first and second derivatives of the two functions, using the results in Table M.10-1. (a) (b) Function y = f ( x ) y = x 2 – 4 x + 10 y = 2 + 4 x x 2 First derivative y ’ = f ’( x ) = d d y x y ’ = 2 x – 4 y ’ = 4 – 2 x Second derivative y ” = f ”( x ) = d d 2 x y 2 y ” = 2 > 0 y ” = –2 < 0 Recall that the first derivative, dy/dx , gives the slope of a tangent to the function at any point on the graph. Note that for both graphs, the value of the first derivative when x = 2 is zero: for (a), dy/dx = 2 x – 4 = 2(2) – 4 = 0, while for (b), dy/dx = 4 – 2 x = 4 – 2(2) = 0. Alternatively put, the tangent to both functions at (2, 6) is horizontal. This is a necessary condition for a point to be a (local) maximum or minimum. If the slope were negative at that point, the function would have a higher value just to the left (and so the point could MATH MODULE 10: CALCULUS RESULTS FOR THE NON-CALCULUS SPEAKER M10-5 O I II (a) (b) x y = x 2 – 4 x + 10 y 6 10 2 O III IV x y = 2 + 4 x x 2 y 6 2 2 FIGURE M.10-2 TABLE M.10-2 Derivatives of the Func- tions in Figure M.10-2
not be a maximum) and a lower value just to the right (and so it could not be a mini- mum). Similarly, if the slope were positive at that point, the function would have a high- er value just to the right (and so the point could not be a maximum) and a lower value just to the left (and so it could not be a minimum). This necessary condition, however, is not sufficient to ensure that we have either a maximum or a minimum! The sign of the second derivative at a point where the first derivative is zero is what tells us whether the point is a maximum or a minimum (or neither). Suppose we were to graph the first derivatives or slopes in Figure M.10-2 (a) and (b) as functions of x . In (a), we notice that to the left of x = 2 (in Zone I), the slope is negative, but to the right of x = 2 (in Zone II), it is positive. In other words, the slope is steadily increasing in value as x increases: the “slope of the slope” is positive . But the second derivative is precisely “the slope of the slope.” As Table M.10-2 shows, the sec- ond derivative of the function in (a) is a constant +2 > 0. If the second derivative of a function is positive at a point where the first derivative is zero, then at this point, the function has a (local) minimum .

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