Suppose that f(2) = -3, g(2) = 2, f '(2) = -4, and g'(2) = 1. Find h'(2). (a) h(x) = 2f(x) - 3g(x) h'(2) = (b) h(x) = f(x)g(x) h'(2) = (c) h(x) = f(x) / g(x) h'(2) = (d) h(x) = g(x) / (1 + f(x)) h'(2) =
Added by Bradley S.
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Using the linearity of the derivative, we have: h'(x) = 2f'(x) + 3g'(x) Now, we can plug in the given values for f'(2) and g'(2): h'(2) = 2(4) + 3(1) = 8 + 3 = 11 So, h'(2) = \boxed{11} for part (a). Show more…
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