By Sterling K. Berberian

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7 If f(x) = x 2 + 3 and g(x) = 2x + 1, then The derivative of a composite function may also be obtained with the following role: Rule 13. The Chain Rule: D x (f(g(x))) = f’(g(x))g’(x) If f is called the outer function and g is called the inner function, then D x (f(g(x))) is the product of the derivative of the outer function [evaluated at g(x)] and the derivative of the inner function. 7, f’(x) = 2x. Therefore, f’(g(x)) = 2g(x) and g’(x) = 2. Hence by the chain rule, Schaum’s Easy Outlines Calculus Page 24 You Need To Know An alternative formulation of the chain rule is the following: Write y = f(u) and u = g(x).

A) Here V = pr 2 h = 64 in 3 , and A = 2prh + pr 2 . To express A as a function of one variable, we solve for h in the first relation (because it is easier) and substitute in the second, obtaining and the critical value is . Then . Thus, in. Now dA/dr > 0 to the right of the critical value, and dA/dr < 0 to the left of the critical value. So, by the first-derivative test, we have a Schaum’s Easy Outlines Calculus Page 47 relative minimum. Since there is no other critical value, that relative minimum is an absolute minimum.

Given this, we seek to determine the maximum of A = defined over the interval f(x) = x (200 - 2x) for x in the interval [0, 100]. In this case, A = f(x) = 200x - 2x 2 We take the derivative of A: f’ (x)=200 - 4x Setting f’(x) = 0 200 - 4x = 0 4x = 200 x = 50 feet Substituting this value into Eq. (3), y = 200 - 2(50) = 200 - 100 = 100 feet Thus, the dimensions of the garden that yield the maximum possible area are x = 50 feet and y = 100 feet. 5 Given a square piece of cardboard with sides equal to 16 inches, Laura would like to construct a box by cutting out four squares, one from each comer.