#$&* course MTH 271 3.31.12 11:45 AM 017. `query 17
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Given Solution: `a The first function you evaluate is x^2 - 3x + 3. You then cube this result. So the breakdown to get f(g(x)) form is f(z) = z^3 g(x) = x^2 - 3x + 3. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 1e 7th edition 2.5.8 inner, outer fns for (x+1)^-.5 Your solution: f(z) = z^ -0.5 and z(x) = x + 1 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 2b 7th edition 2.5.24 der of f(t) = (9t+2)^(2/3) by gen power rule Your solution: The rule for the derivative of this function is f(x) = x^n à f’(x) = n * (x^n-1) * x’ So, using this rule, the derivative is f’(t) = (2/3)*((9t + 2)^-1/3) * 9 à F’(t) = 6(9t + 2)^-1/3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a This function is of the form u^(2/3), where u = 9 t + 2. The derivative of f(t) = u^p is f ' (t) = p * u^(p-1) * u ', by the general power rule. Here p = 2/3, and u ' = (9t + 2)' = 9 so we have f ' (t) = p u^(p-1) * u ' = 2/3 * u^(2/3 - 1) * 9 = 6 * u^(-1/3), or since u = 9 t + 2 f ' (t) = 6 ( 9 t + 2)^(-1/3). ** STUDENT COMMENT: This explanation loses me. Is there any way you could explain it different. I am having a hard time following exactly where I messed up. INSTRUCTOR RESPONSE: The power function rule (x^n) ' = n x ^ (n - 1) could be written just as well using u as the variable; it would read (u^n) ' = n u ^ (n - 1), where the ' now means the derivative with respect to u. Thus (u^(2/3)) ' = 2/3 u^(2/3 - 1) = 2/3 u^(-1/3). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 2c 7th edition 2.5.28 der of f(x) = (25+x^2)^(-1/2) by gen power rule Your solution: This function can be written as two partial functions: F(z) = z^-0.5 and z(x) =25 + x^2 The derivatives are: f’(z) = -(1/2) *(z^-3/2) and z’(x) = 2x Combined, the full derivative is f’(x) = (-1/2) * (25 + x^2)^(-3/2) * 2x à F’(x) = - x *(25 + x^2)^ (-3/2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): OK ------------------------------------------------ Self-critique Rating: 3" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!