open qa 161

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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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Your solution:

2xy + x^2 y’

confidence rating #$&*:2

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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 ' .

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Self-critique (if necessary):

***the way this expression is written it also could be read as x^(2y)

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Self-critique Rating:

@& The expression could not be read as x^(2 y). Exponentiation precedes multiplication. So x is raised to the power 2 before anything is multiplied.

Your solution, of course, was fine.*@

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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?

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Your solution:

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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.

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Self-critique (if necessary):

I wasn’t able to come up with this one on my own… It sounded too confusing.

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Question: `q003. If y is a function of x, then what is the derivative of the expression x^2 y^3?

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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.

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Self-critique (if necessary):

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Question: `q004. The equation 2x^2 y + 7 x = 9 can easily be solved for y. What is the result?

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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 ).

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Self-critique (if necessary):

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Question: `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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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.

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Self-critique (if necessary):

Had to peak in the beginning of the solution to get me started… but other than that it was fine.

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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 '.

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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.

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Self-critique (if necessary):

When we solve for y’ this way, what exactly IS y’???

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Self-critique Rating:

@& y ' is the derivative of y with respect to x. Interpreted as the instantaneous rate of change of y with respect to x.*@

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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.

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Your solution:

2(2xy^3 + 3y^2x^2) - 3(y^2 + 2xy) = 0

4xy^3 + 6y^2x^2 - 3y^2 - 6xy = 0

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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... .

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Self-critique (if necessary):

### I just DO NOT see where the y’ in this solution comes from??? I know I was supposed to have it in my solution somewhere…. But I am still not sure how this needs to be done.

4 x y^3 + 6 x^2 y^2 y' - 3 y^2 - 6xy y' = 0.

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Self-critique Rating:ok

@& The derivatives are all taken with respect to x. So the derivative of x^2 is just 2 x.

However the expression y^2 is the square of y, but y is itself a function of x. So y^2 is a composite function, z = y^2 where y = y(x), whatever y(x) happens to be.

The derivative of this function is 2 y * y ' (x), by the chain rule.*@

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#*&!

&#This looks good. See my notes. Let me know if you have any questions. &#