course Mth 271 017. `query 17
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q 2.5.8 inner, outer fns for (x+1)^-.5 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q 2.5.20 der of f(t) = (9t+2)^(2/3) by gen power rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: This function is of the form u^(2/3), where u = 9 t + 2. f(t) = u^p is f ' (t) = p * u^(p-1) * u ', 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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Given Solution: `a 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). ** STUDENT COMMENT: This explanation loses me. Is there any way you could explain it different. I am having a hard time following exactly where I messed up. INSTRUCTOR RESPONSE: The power function rule (x^n) ' = n x ^ (n - 1) could be written just as well using u as the variable; it would read (u^n) ' = n u ^ (n - 1), where the ' now means the derivative with respect to u. Thus (u^(2/3)) ' = 2/3 u^(2/3 - 1) = 2/3 u^(-1/3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok ********************************************* Question: `q 2.5.28 der of f(x) = (25+x^2)^(-1/2) by gen power rule YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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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Given Solution: `a 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). ** STUDENT QUESTION f(x) = (25 + x^2)^-1/2 f(x) = -1/2 (25 + x^2) (25 + x^2)' f(x) = (-12.5 - 1/2x^2) (2x) Is what I had for my answer. I still don't quite understand where I am messing up. Although, this section I am having trouble with a little bit. INSTRUCTOR RESPONSE you appear to occasionally overlook the power function rule in this case you have (u^(-1/2)) ' = -1/2 u^(-1/2 - 1) = -1/2 u^(-3/2) So your expression f(x) = -1/2 (25 + x^2) (25 + x^2)' should have read f(x) = -1/2 (25 + x^2)^(-3/2) * (25 + x^2)' That appears to be the only step you're missing, so it's clear you are on the right track. I'm confident you'll clear this up, but be sure to let me know if not. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating:ok "