course MTH 271 10/26, 1700 assignment #016016. Implicit differentiation.
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15:47:25 `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) + (x^2)(y') confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:47:35 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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15:50:50 `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)(y') confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:51:08 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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15:52:23 `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 --> (2x)(y^3) +(x^2)(3y^2) 2xy^3 + 3(x^2)(y^2)y' confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:52:48 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 --> ------------------------------------------------ Self-critique Rating:ent: 3
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15:54:37 `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result?
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RESPONSE --> 2x^2 y + 7x = 9 2x^2 y = 9 - 7x y=(9-7x) / (2x^2) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:54:47 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 Rating:ent: 3
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15:58:29 `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 x^2 ) - 7 / ( 2 x ) When x=1, y=1. y'= (0)(2x^2) - 9(4x) - (0(2x) - 7(2)) y'=-36x +14 y'(1)=-36(1) + 14 =-22 confidence rating: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:58:51 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 --> While I correctly figured the y value, my value for the slope or y' was incorrect. I now see that this was b/c I did not find a common denominator for the two quantities. ------------------------------------------------ Self-critique Rating:ent: 3
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15:59:34 `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 --> 2(xy + x^2y') + 7=0 2xy + 2x^2y'=-7 2x^2y'=-2xy - 7 y'=(-2xy - 7)/(2x^2) y'when x=1, y=1 y'=(-2(1)(1) - 7) / (2(1^2)) y'=(-2 - 7) / 2 y'=-9 / 2 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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15:59:53 `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 --> 2 x^2 y^3 - 3x y^2 - 4=0 2x^2y^3 - 3xy^2 -4=0 (4x)(y^3) + (2x^2)(3y^2) - (3)(y^2) - (3x)(2y)=0 4xy^3 + 6x^2y^2y' - 3y^2 -6xyy'=0 6x^2y^2y' - 6xyy'= -4xy^3 + 3y^2 y'(6x^2y^2 -6xy) = -4xy^3 + 3y^2 y'=(-4xy^2 + 3y^2)/(6x^2y^2 - 6xy) y'=(-4(1)(2^2) + 3(2^2)) / (6(1^2)(2^2) - (6(1)(2))) y'=(-8 + 12) / (24 -12) y'=(4)/(12) y'= 1/3 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:00:07 `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 --> 4xy^3 + 6x^2y^2y' - 3y^2 -6xyy'=0 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:04:30 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' - 6xyy' = -4xy^3 - 3y^2 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:05:46 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 - 6xy)=-4xy^3 + 3y^2 confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:06:46 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 = (-4xy^3 + 3y^2) / y' confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:09:24 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( (-4xy^2 + 3y) / (6x^2y - 6x) ) confidence rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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16:10:09 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 rating: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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