course mth 173 assignment #016016. Implicit differentiation.
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21:04:39 `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 --> If y is in respect to x then we can say y(x) = x^2 y using the chain rule we get y'= (x^2)' (y) + (y)'(x^2) = (2x)(y) + (y')(x^2) =2xy + x^2 y' confidence assessment: 2
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21:05:14 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 --> okay self critique assessment: 3
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21:08:07 `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 --> y(x)= y(x)^3 y'(x)= f'(y(x)) (y'(x)) = (3y^2)(y') confidence assessment: 2
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21:10:18 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 --> okay self critique assessment: 3
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21:14:48 `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 --> y(x)= x^2 y^3 y'(x)= (x^2)' (y^3) + (y^3)' (x^2) = (2x)(y^3) + (3y^2)(x^2) = 2xy^3 + 3y^2 x^2 confidence assessment: 2
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21:17:14 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 didn't add the final y' but I will remember to do that in the future self critique assessment: 2
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21:18:56 `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 assessment: 2
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21:19:28 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 --> okay self critique assessment: 3
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21:25:27 `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 x^2 ) - 7 / ( 2 x ) y(1)= 1 y= (9 * (2x^2)^-1) - (7(2x)^-1) y'= (9 * (2x^2)^-1)' - (7(2x)^-1)' =(9 * -(4x)^-2) - (7 -(2^-2)) (9 * -(4x)^-2) + 1.75 When y' is evaluated for x=1 we get 1.1875 confidence assessment: 1
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21:28:15 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 --> I got the y value correct because it can be plugged right in and is fairly simple, however the harder part of finding the derivative is where I messed up, instead of using the quotient rule I used the product rule but I think that I may have messed something in the process of finding the direvative, therefore making my answer wrong. self critique assessment: 2
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21:32:54 `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((x^2)' y +x^2 y') +7 = 0 2((2x) y + x^2 y') +7 = 0 2(2xy + x^2y') +7 = 0 4xy + 2x^2 y' +7 = 0 4xy + 2x^2y' = -7 2x^2 y'= -7 -4xy y'= (-7-4xy) / 2y^2 y'= -5.5 which is the same as the right answer in the last question. confidence assessment: 3
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21:34:30 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 --> okay self critique assessment: 3
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21:42:43 `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 - 3 x y^2 - 4 = 0 when x=1 and y=2 then the equation does equal zero. 2 x^2 y^3 - 3 x y^2 - 4 = 0 (2x^2)'(y^3) + (y^3)'(2x^2) - (3x)'(y^2) + (y^2)'(3x)=0 4xy^3 + 6y^2 x^2 - 3y^2 + 2y * 3x= 0 4xy^3 + 6y^2 x^2 y' - 3y^2 + 6xy y'=0 y'(6y^2 x^2 -6xy) = -4xy^3 + 3y^2 y'= (-4xy^3 + 3y^2) / 6y^2 x^2 -6xy) y'= -1.67 confidence assessment: 2
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21:44:20 `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 --> I'm going forward to see whats after ""or"", hopefully it's on the next page! But I look good so far. confidence assessment: 2
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21:44:34 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 --> okay confidence assessment: 3
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21:44:58 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 --> okay confidence assessment: 3
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21:45:10 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 --> okay3 confidence assessment: 3
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21:46:04 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 --> I guess I could have simplified one more step but up untill this point my answer is the same confidence assessment: 2
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21:46:24 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 --> okay confidence assessment: 3
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21:55:00 `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 --> x^2 sin (y) - sin(xy) = 0 For x= 3 and y= 'pi (x^2)'(sin(y)) + (sin(y))'(x^2) - (sin(xy))' = 0 (2x)(sin(y)) + (cos(y))(x^2) - (cos(xy))' =0 2x sin(y) + cos(y) x^2 y' - cos(xy) y'= 0 y'(cos(y)x^2 - cos(xy)) = -2x sin(y) y'= (-2x sin(y)) / (cos(y)x^2 - cos(xy)) y'= 0 confidence assessment: 2
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21:59:06 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 --> I messed up in find the derivative of sin(xy) because I didn't preceed the derivative with x and y as seen here: y cos(xy) - x cos(xy which messed up my answer., re working the problem i got the same answer as you. self critique assessment: 2
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