#$&* course Mth 173 3/18 10:56 PM 016. Implicit differentiation.
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Given Solution: `aBy the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' . STUDENT QUESTION I understand this concept except for where the 2xy came from? ------------------------------------------------ Self-critique Rating:ent: 1 INSTRUCTOR RESPONSE (x^2 y) ' = (x^2) ' y + x^2 * y ' The 2 x y comes from (x^2) ' y: (x^2) ' = 2x so (x^2)' y = 2 x y. Thus (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): where it was why I didnt view it in terms of the rules. so f(x)= x^2 g(x)= y f'(x)g(x) + f(x)g'(x) 2xy + x^2y' ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q002. Note that if y is a function of x, then y^3 is a composite function, evaluated from a given value of x by the chain of calculations that starts with x, then follows the rule for y to get a value, then cubes this value to get y^3. To find the derivative of y^3 we would then need to apply the chain rule. If we let y' stand for the derivative of y with respect to x, what would be the derivative of the expression y^3 with respect to x? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: chain rule is the one inside another right? so that must mean f(g(x)) so f(g(y^3)) cant say I understand the question well g(x)= y^3 f(x)= y g'(x) * f'(g(x) ) 3y^2 * y' confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): so is the reason for the y' being there beause we have something to the essence of y=x * y seems kind of recursive. but so we also have to include it once for the f(x), and then for the right side of the equal sign. ------------------------------------------------ Self-critique 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 '. STUDENT QUESTION I am confused about the last part of this problem why is g` not just 3y^2, I do not understand [ y ' * 3 y^2 ] why is this multiplied by y`? I see were the 3y^2 originated from but why is there a y`? INSTRUCTOR RESPONSE y is a function of x. So y^3 is a composite function, as shown in the preceding question. Your answer to that question was that the derivative of (y(x))^3 was y`(x)*3*(y(x))^2. Abbreviated, your result is y ' * 3 y^2, the same as 3 y^2 y '. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Still a little iffy on the reasoning of when we take the derivative of f(x)=y that we must also include another y' ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 2x^2y + 7x = 9 ???for some reason just thought, we have to do PEMDAS (parenthesis exponent etc) backwards when doing across equal sign 2x^2y= 9 - 7x y= (9-7x)/2x^2 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aStarting with 2 x^2 y + 7 x = 9 we first subtract seven x from both sides to obtain 2 x^2 y = 9 - 7 x. We then divide both sides by 2 x^2 to obtain y = (9 - 7 x ) / (2 x^2), or if we prefer y = 9 / (2 x^2 ) - 7 / ( 2 x ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `q005. Using the form y = 9 / (2 x^2 ) - 7 / ( 2 x ), what is y and what is y ' when x = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y(1)= 1 ???exactly what are the steps for the derivatve, im just looking at is as keeping everything as is, and just changing x^2 to 2x, and changing 2x to 2 afraid might be skipping a quotient or product rule though y'(1)= 9/2 - 7/2 confidence rating #$&*:1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): There is something I am not seeing about this problem maybe where they are in the denominator, thats why i should reverse the signs, still dont see how the denominator under -9 became 1. ------------------------------------------------ Self-critique Rating:
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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:OK ********************************************* 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): 2 x^2 y^3 - 3 x y^2 - 4 = 0. 4xy^3 + 2x^2 * 3y^2 * y' - 3y^2 -3x * 2y * y' = 0 2x^2 * 3y^2 * y' - 3y^2 - 3x * 2y * y' = -4xy^3 2x^2 * 3y^2 * y'- 3x * 2y * y' = 3y^2 -4xy^3 6x^2y^2y' - 6xy * y' = 3y^2 -4xy^3 y'(6x^2y^2 - 6xy)=3y^2 -4xy^3 y'=(3y^2 -4xy^3)/(6x^2y^2 - 6xy) y'=(3y -4xy^2) / (6x^2y - 6x) 2 * 1 * 8* - 3 * 1 * 4 - 4 = 0 16 -12 -4 = 0 16 - 16 = 0 0 = 0 checks out y'=(3 * 2 - 4 * 1 * 4) / (6 * 1 * 2 -6 * 1) y'=(6 - 16) / (12 - 6) y'=-10/6 ------------------------------------------------ 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... . STUDENT QUESTION I see what was done in the solution once I read that I did not have to solve for y. My approach to this was to solve for y and that did not turn out well. My question is how do I know when to solve for y and when not to? INSTRUCTOR RESPONSE Once you're used to implicit differentiation, unless otherwise stated, then you can make that decision for yourself. Of course for most expressions you won't be able to solve for y, so the decision is already made by the problem. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* 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. 2xsin(y) + x^2cos(y)y' ???I am not sure how to take the derivative of sin(xy), i keep getting the wrong result confidence rating #$&*:0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. STUDENT QUESTION I think that the derivative of sin (xy) is cos (xy) but in the solution ( y + x y ' ) cos(xy) is the derivitive. I had this same problem on a earlier problem also, can you maybe explain to me were I am messing this up? Or were the y+xy` came from? INSTRUCTOR RESPONSE The derivative of sin(xy) is (xy)' * cos(xy), by the chain rule. (xy)' = x ' y + x y ' = 1 * y + x y ' = y + x y '. Thus the derivative of sin(xy) is (y + x y ' ) cos(xy). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I am still having trouble with the The derivative of sin(xy) is (xy)' * cos(xy), by the chain rule. (xy)' = x ' y + x y ' = 1 * y + x y ' = y + x y '. Thus the derivative of sin(xy) is (y + x y ' ) cos(xy). isnt the chain rule f(g(x)) = f'(g(x)) g'(x) g(x)= xy f(x)= sin(xy) f'(g(x))= cos(xy) g'(x)= y' ------------------------------------------------ Self-critique Rating:"