course Math 271 07-28-2008½¨ù©ïz’´}yðéèœÔÉîÔ׸Èçz·
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14:58:09 `q001. Suppose that y is a function of x. Then the derivative of y with respect to x is y '. The derivative of x itself, with respect to x, is by the power function rule 1, x being x^1. What therefore would be the derivative of the expression x^2 y?
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RESPONSE --> 2x(y) confidence assessment: 1
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14:59:03 By the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' .
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RESPONSE --> Ok self critique assessment: 3
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15:02:51 `q002. Note that if y is a function of x, then y^3 is a composite function, evaluated from a given value of x by the chain of calculations that starts with x, then follows the rule for y to get a value, then cubes this value to get y^3. To find the derivative of y^3 we would then need to apply the chain rule. If we let y' stand for the derivative of y with respect to x, what would be the derivative of the expression y^3 with respect to x?
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RESPONSE --> 3y^2 + 3x^2 +xy confidence assessment: 1
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15:06:44 The derivative of y^3 with respect to x can be expressed more explicitly by writing y^3 as y(x)^3. This reinforces the idea that y is a function of x and that we have a composite f(y(x)) of the function f(z) = z^3 with the function y(x). The chain rule tells us that the derivative of this function must be (f ( y(x) ) )' = y ' (x) * f ' (y(x)), in this case with f ' (z) = (z^3) ' = 3 z^2. The chain rule thus gives us (y(x)^3) ' = y ' (x) * f ' (y(x) ) = y ' (x) * 3 * ( y(x) ) ^2. In shorthand notation, (y^3) ' = y ' * 3 y^2. This shows how the y ' comes about in implicit differentiation.
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RESPONSE --> I don't understand when to use the f(z) function and when not to. Also Why do we lose the x in the final answer? self critique assessment: 2
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15:22:28 `q003. If y is a function of x, then what is the derivative of the expression x^2 y^3?
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RESPONSE --> 3y^2 * 2x [y(x)^3] confidence assessment: 2
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15:24:07 The derivative of x^2 y^3, with respect to x, is (x^2 y^3) ' = (x^2)' * y^3 + x^2 * (y^3) ' = 2x * y^3 + x^2 * [ y ' * 3 y^2 ] = 2 x y^3 + 3 x^2 y^2 y '. Note that when the expression is simplified, the y ' is conventionally placed at the end of any term of which is a factor.
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RESPONSE --> I am obviously using the wrong rules here! self critique assessment: 3
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15:26:02 `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result?
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RESPONSE --> y=-7x+9 /2x^2 confidence assessment: 3
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15:26:25 Starting with 2 x^2 y + 7 x = 9 we first subtract seven x from both sides to obtain 2 x^2 y = 9 - 7 x. We then divide both sides by 2 x^2 to obtain y = (9 - 7 x ) / (2 x^2), or if we prefer y = 9 / (2 x^2 ) - 7 / ( 2 x ).
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RESPONSE --> self critique assessment: 3
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15:30:27 `q005. Using the form y = 9 / (2 x^2 ) - 7 / ( 2 x ), what is y and what is y ' when x = 1?
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RESPONSE --> y=9/2 - 7/2 = 1 y'=9/4x - 7/2x = 9/4 - 7/2 = -5/4 confidence assessment: 3
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15:31:11 y ' = 9/2 * ( 1/x^2) ' - 7/2 ( 1 / x) ' = 9/2 * (-2 / x^3) - 7/2 * (-1 / x^2) = -9 / x^3 + 7 / (2 x^2). So when x = 1 we have y = 9 / (2 * 1^2 ) - 7 / ( 2 * 1 ) = 9/2 - 7/2 = 1 and y ' = -9 / 1^3 + 7 / (2 * 1^2) = -9 + 7/2 = -11 / 2.
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RESPONSE --> self critique assessment: 3
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15:35:53 `q006. The preceding calculation could have been obtained without solving the original equation 2x^2 y + 7 x = 9 for y. We could have taken the derivative of both sides of the equation to get 2 ( (x^2) ' y + x^2 y ' ) + 7 = 0, or 2 ( 2x y + x^2 y ' ) + 7 = 0. Complete the simplification of this equation, then solve for y ' . Note that when x = 1 we have y = 1. Substitute x = 1, y = 1 into the equation for y ' and see what you get for y '.
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RESPONSE --> y=1x-7/2 y' y'=-3 confidence assessment: 2
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15:40:40 Starting with 2 ( 2x y + x^2 y ' ) + 7 = 0 we divide both sides by 2 to obtain 2x y + x^2 y ' + 7/2 = 0. Subtracting 2 x y + 7 / 2 from both sides we obtain x^2 y' = - 2 x y - 7 / 2. Dividing both sides by x^2 we end up with y ' = (- 2 x y - 7 / 2) / x^2 = -2 y / x - 7 / ( 2 x^2). Substituting x = 1, y = 1 we obtain y ' = -2 * 1 / 1 - 7 / ( 2 * 1^2) = - 11 / 2. Note that this agrees with the result y ' = -11/2 obtained when we solved the original equation and took its derivative. This shows that it is not always necessary to explicitly solve the original equation for y. This is a good thing because it is not always easy, in fact not always possible, to solve a given equation for y.
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RESPONSE --> Ok I understand where I messed up. self critique assessment: 2
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16:10:35 `q007. Follow the procedure of the preceding problem to determine the value of y ' when x = 1 and y = 2, for the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0. Also validate the fact that x = 1, y = 2 is a solution to the equation, because if this isn't a solution it makes no sense to ask a question about the equation for these values of x and y.
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RESPONSE --> I don't know if I am supposed to find the derivative for this first or not, and if so I can't simplify it to find the answer? confidence assessment: 1
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16:13:04 `qNote that this time we do not need to solve the equation explicitly for y. This is a good thing because this equation is cubic in y, and while there is a formula (rather a set of formulas) to do this it is a lengthy and messy process. The derivative of the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 is (2 x^2 y^3) ' - (3 x y^2) ' - (4)' = 0. Noting that 4 is a constant, we see that (4)' = 0. The derivative of the equation therefore becomes (2 x^2) ' * y^3 + 2 x^2 * ( y^3) ' - (3 x ) ' * y^2 - 3 x * (y ^2 ) ' = 0, or
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RESPONSE --> confidence assessment: 1
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16:15:10 4 x y^3 + 6 x^2 y^2 y' - 3 y^2 - 6xy y' = 0. Subtracting from both sides the terms which do not contain y ' we get
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RESPONSE --> 6x^2y^2y' - 6xy y' = -4xy^3 - 3y^2 confidence assessment: 2
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16:16:27 6 x^2 y^2 y ' - 6 x y y ' = - 4 x * y^3 + 3 y^2. Factoring out the y ' on the left-hand side we have
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RESPONSE --> y'(6x^2y^2 0 6xy)= -4xy^3 + 3y^2 confidence assessment: 3
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16:18:01 y ' ( 6 x^2 y^2 - 6 x y) = -4 x y^3 + 3 y^2. Dividing both sides by the coefficient of y ':
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RESPONSE --> (6x^2y^2 - 6xy) / y' = -4xy^3 + 3y^2 / y' confidence assessment: 1
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16:19:54 y ' = (- 4 x * y^3 + 3 y^2) / ( 6 x^2 y^2 - 6 x y ) . The numerator and denominator have common factor y so we end up with
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RESPONSE --> y'= y^2 * [ (-4xy^3 + 3) / (6x^2- 6xy) confidence assessment: 2
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16:22:03 y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ). Now we see that if x = 1 and y = 2 the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0 gives us 2 * 1^2 * 2^3 - 3 * 1 * 2^2 - 4 = 0, or 16 - 12 - 4 = 0, which is true. Substituting x = 1 and y = 2 into the equation y' = (- 4 x * y^2 + 3 y) / ( 6 x^2 y - 6 x ) we get y ' = (- 4 * 1 * 2^2 + 3 * 2) / ( 6 * 1^2 * 2 - 6 * 1 ) = (-16 + 6) / (12 - 6) = -10 / 6 = -5/3 or -1.66... .
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RESPONSE --> confidence assessment: 3
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16:22:17 `q008. (Mth 173 only; Mth 271 doesn't require the use of sine and cosine functions). Follow the procedure of the preceding problem to determine the value of y ' when x = 3 and y = `pi, for the equation x^2 sin (y) - sin(xy) = 0.
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RESPONSE --> confidence assessment: 3
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16:22:21 Taking the derivative of both sides of the equation we obtain (x^2) ' sin(y) + x^2 (sin(y) ) ' - ( sin (xy) ) '. By the Chain Rule (sin(y)) ' = y ' cos(y) and (sin(xy)) ' = (xy) ' cos(xy) = ( x ' y + x y ' ) cos(xy) = (y + x y ' ) cos(xy). So the derivative of the equation becomes 2 x sin(y) + x^2 ( y ' cos(y)) - ( y + x y ' ) cos(xy) = 0. Expanding all terms we get 2 x sin(y) + x^2 cos(y) y ' - y cos(xy) - x cos(xy) y ' = 0. Leaving the y ' terms on the left and factoring ou y ' gives us [ x^2 cos(y) - x cos(xy) ] y ' = y cos(xy) - 2x sin(y), so that y ' = [ y cos(xy) - 2x sin(y) ] / [ x^2 cos(y) - x cos(xy) ]. Now we can substitute x = 3 and y = `pi to get y ' = [ `pi cos( 3 * `pi) - 2 * 3 sin(`pi) ] / [ 3^2 cos(`pi) - 3 cos(3 * `pi) ] = [ `pi * -1 - 2 * 3 * 0 ] / [ 9 * -1 - 3 * -1 ] = -`pi / 6.
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RESPONSE --> self critique assessment: 3
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