course MTH 173 11JULY2010 AT 2028 016. Implicit differentiation.
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Given Solution: `aBy the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' . STUDENT QUESTION I understand this concept except for where the 2xy came from?
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Given Solution: `aThe 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q003. If y is a function of x, then what is the derivative of the expression x^2 y^3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 y^3= (x^2*y^3)’= (2x+y^3)+ [(x^2)(3y^2)y’] confidence rating #$&* ok ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe 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. STUDENT QUESTION Ok I understand everything until the last part of the answer. How come the 3 only got combined with the x^2? INSTRUCTOR RESPONSE a * b * c = c * b * a = b * c * a = etc.. The order in which the quantities are listed in a string of multiplications doesn't matter. So x^2 * [ y ' * 3 y^2 ] means the same thing as 3 y^2 * x^2 * y ' which means the same as 3 x^2 y^2 y '. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x^2 y + 7 x = 9 solve for y 2x^2 y + 7 x = 9== (2x^2)y=9-7x y=(9-7x)/(2x^2) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aStarting 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 ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q005. Using the form y = 9 / (2 x^2 ) - 7 / ( 2 x ), what is y and what is y ' when x = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 9 / (2 x^2 ) - 7 / ( 2 x ) y’= 9/2 * (-2 / x^3) - 7/2 * (-1 / x^2) y’= -9 / x^3 + 7 / (2 x^2) y(1)=9/2(1^2) -7 / 2(1)=1 y’(1)= -9 / (1)^3 + 7 / [2 (1^2)]= -18/2 + 7/2 =-11/2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `ay ' = 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q006. The preceding calculation could have been obtained without solving the original equation 2x^2 y + 7 x = 9 for y. We could have taken the derivative of both sides of the equation to get 2 ( (x^2) ' y + x^2 y ' ) + 7 = 0, or 2 ( 2x y + x^2 y ' ) + 7 = 0. Complete the simplification of this equation, then solve for y ' . Note that when x = 1 we have y = 1. Substitute x = 1, y = 1 into the equation for y ' and see what you get for y '. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2 ( 2x y + x^2 y ' ) + 7 = 0== 2(2xy+x^2y’)=-7== subtract 7 from both sides (2xy+x^2y’)/2=-7/2 divided both sides by 2 2xy+x^2y’=-7/2 ==subtract 2xy from both sides x^2y’=-7/2 -2xy== divide both sides by x^2 y’=[-7/2 -2xy]/x^2=-2y/x-7/(2x^2) y’= -2(1/1)-7/[2x(1^2)]= -11/2 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aStarting 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q007. Follow the procedure of the preceding problem to determine the value of y ' when x = 1 and y = 2, for the equation 2 x^2 y^3 - 3 x y^2 - 4 = 0. Also validate the fact that x = 1, y = 2 is a solution to the equation, because if this isn't a solution it makes no sense to ask a question about the equation for these values of x and y. Your Solution: 2 x^2 y^3 - 3 x y^2 - 4 = 0== 2(1)(2^3)-3(1)(2^2)-4=16-12-4=0 4xy^3+6x^2y^2y’-6xyy’-3y^2=0 4xy^3-3y^2=-6x^2y^2y’+6xyy’==got y’ to one side 4xy^3-3y^2=(-6x^2y^2+6xy)y’==pulled out y’ on right side (4xy^3-3y^2)/ (-6x^2y^2+6xy) = y’== divided both sides by (-6x^2y^2+6xy) (4xy^2-3y)/ (-6x^2y+6x) = y’==reduce by y when x=1 and y=2==y’==((4(1)4)-6)/ (-6(1)2+6)=10/-6 &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique rating #$&* ********************************************* Question: `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 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 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 y ' ( 6 x^2 y^2 - 6 x y) = -4 x y^3 + 3 y^2. Dividing both sides by the coefficient of y ': 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 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... . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):ok ------------------------------------------------ Self-critique rating #$&*3 ********************************************* Question: `q008. (Mth 173 only; Mth 271 doesn't require the use of sine and cosine functions). Follow the procedure of the preceding problem to determine the value of y ' when x = 3 and y = `pi, for the equation x^2 sin (y) - sin(xy) = 0. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 sin (y) - sin(xy) = 0 == 9sin(pi) –sin(3pi)==0 2xsin(y) + x^2cos(y)y’ – cos(xy) (y+xy’)== distribute cos(xy) 2xsin(y) + x^2cos(y)y’ – ycos(xy) + xcos(xy)y’=0 get y’ together on one side 2xsin(y) – ycos(xy) = -x^2cos(y)y’- xcos(xy)y’== factor out y’ on right side 2xsin(y) – ycos(xy) = [-x^2cos(y)- xcos(xy)]y’==divide both sides by[-x^2cos(y)- xcos(xy)] y’=2xsin(y) – ycos(xy) / [-x^2cos(y)- xcos(xy)] y’=2(3)sin(pi) – picos(3pi) / [-3^2cos(pi)- 3cos(3pi)]= subsututed x=3 and y=pi y’=6sinpi-picos(3pi)/ [9cos(pi)- 3cos(3pi)]==approx. -.5236 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aTaking 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&*3 "