#$&* course Mth 271 Question: `q001. Suppose that y is a function of x. Then the derivative of y with respect to x is y '. The derivative of x itself, with respect to x, is by the power function rule 1, x being x^1. What therefore would be the derivative of the expression x^2 y?
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Given Solution: `aBy the product rule we have (x^2 y) ' = (x^2) ' y + x^2 * y ' = 2 x y + x^2 y ' . &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ***the way this expression is written it also could be read as x^(2y) ------------------------------------------------ Self-critique Rating:
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Given Solution: `aThe derivative of y^3 with respect to x can be expressed more explicitly by writing y^3 as y(x)^3. This reinforces the idea that y is a function of x and that we have a composite f(y(x)) of the function f(z) = z^3 with the function y(x). The chain rule tells us that the derivative of this function must be (f ( y(x) ) )' = y ' (x) * f ' (y(x)), in this case with f ' (z) = (z^3) ' = 3 z^2. The chain rule thus gives us (y(x)^3) ' = y ' (x) * f ' (y(x) ) = y ' (x) * 3 * ( y(x) ) ^2. In shorthand notation, (y^3) ' = y ' * 3 y^2. This shows how the y ' comes about in implicit differentiation. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t able to come up with this one on my own… It sounded too confusing. ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q003. If y is a function of x, then what is the derivative of the expression x^2 y^3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: X^2 y^3’ = 2xy^2 + 3y^2y’ confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* 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 = 9 - 7x Y = (9 - 7x)/2x^2 confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aStarting with 2 x^2 y + 7 x = 9 we first subtract seven x from both sides to obtain 2 x^2 y = 9 - 7 x. We then divide both sides by 2 x^2 to obtain y = (9 - 7 x ) / (2 x^2), or if we prefer y = 9 / (2 x^2 ) - 7 / ( 2 x ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q005. Using the form y = 9 / (2 x^2 ) - 7 / ( 2 x ), what is y and what is y ' when x = 1? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y = 9/2 - 7/2 = 1 Y’ = 9/2 * 1/x^2 - 7/2 * 1/x = 9/2 * -2/x^3 - 7/2 * -1/x^2 = -9/x^3 + 7/2x^2 X =1 -9 + 7/2 = -18/2 + 7/2 = -11/2 confidence rating #$&*:2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): Had to peak in the beginning of the solution to get me started… but other than that it was fine. ------------------------------------------------ Self-critique Rating: ********************************************* 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: 4xy + 2x^2 y’ = -7 2x^2 y’ = -7 - 4xy Y’ = (-7 - 4xy)/2x^2 Y =( -7 -4)/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): When we solve for y’ this way, what exactly IS y’??? ------------------------------------------------ Self-critique Rating: