course MTH 271 10/26,1600 assignment #017017. `query 17
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15:29:52 2.5.2 inner, outer fns for (x^2-3x+3)^3
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RESPONSE --> Inner: z=x^2 - 3x + 3 Outer: y=z^3 confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:30:08 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 Rating:ent: 3
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15:30:54 2.5.8 inner, outer fns for (x+1)^-.5
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RESPONSE --> Inner: z=(x + 1) Outer: y=z^(-1/2) confidence rating: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:31:04 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 Rating:ent: 3
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15:32:46 2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule
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RESPONSE --> (2/3)(9t + 2)^(-1/3) * 9 6(9t + 2)^(-1/3) 6/(9t + 2)^(1/3) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:33:44 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 took my solution one step further by placing the (9t + 2)^(-1/3) as the denominator. ------------------------------------------------ Self-critique Rating:ent: 3
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15:37:08 2.5.32 der of f(x) = (25+x^2)^(-1/2) by gen power rule
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RESPONSE --> nu^(n-1)du/dx (-1/2)(25 + x^2)^(-3/2) * (2x) (-x)/(25 + x^2)^(3/2) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:38:37 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 Rating:ent: 3
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