course Mth 271 I thought I had submitted this before, but I must have forgot to. ????????v?????assignment #017
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14:38:53 2.5.2 inner, outer fns for (x^2-3x+3)^3
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RESPONSE --> The inner function is (x^2-3x+3) The outer function is u^3 confidence assessment:
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14:39:20 The first function you evaluate is x^2 - 3x + 3. You then cube this result. So the breakdown to get f(g(x)) form is f(z) = z^3 g(x) = x^2 - 3x + 3. **
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RESPONSE --> self critique assessment: 3
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14:40:34 2.5.8 inner, outer fns for (x+1)^-.5
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RESPONSE --> f(z) = z^-.5 g(x) = x + 1 confidence assessment: 3
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14:40:49 The first function you evaluate is x+1. You then take this result to the -5 power. So the breakdown to get f(g(x)) form is f(z) = z^-.5 g(x) = x+1. **
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RESPONSE --> self critique assessment: 3
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17:12:11 2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule
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RESPONSE --> f(t) = (9t+2)^(2/3) using the general power rule f'(t) = 2/3 * (9t+2) d/dt(9t+2) f'(t) = 2/3 * (9t+2)(9) f'(t) = 6(9t+2) = 54t + 12 confidence assessment: 3
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17:13:37 This function is of the form u^(2/3), where u = 9 t + 2. The derivative of f(t) = u^p is f ' (t) = p * u^(p-1) * u ', by the general power rule. Here p = 2/3, and u ' = (9t + 2)' = 9 so we have f ' (t) = p u^(p-1) * u ' = 2/3 * u^(2/3 - 1) * 9 = 6 * u^(-1/3), or since u = 9 t + 2 f ' (t) = 6 ( 9 t + 2)^(-1/3). **
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RESPONSE --> I forgot that u needed to be raised to the p-1 power. self critique assessment: 2
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17:20:29 2.5.32 der of f(x) = (25+x^2)^(-1/2) by gen power rule
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RESPONSE --> f(x) = (25+x^2)^(-1/2) f(u) = x^(-1/2) u(x) = 25+x^2 f'(x) = (-1/2)(25+x^2)^(-3/2) * (25+x^2)' f'(x) = (-1/2)(25+x^2)^(-3/2) * 2x f'(x) = -x(25+x^2)^(-3/2) f'(x) = -x / (25+x^2)^(3/2) confidence assessment: 3
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17:20:50 Using inner function u = 25+x^2 and outer function = (u)^(-1/2) we get n u^(n-1) du/dx is -1/2 * u^(-3/2) * 2x = - x ( 25+x^2)^(-3/2). **
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RESPONSE --> self critique assessment: 3
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