Assignment 16 QA

course Mth 173

XwHV|yQassignment #016

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016. Implicit differentiation.

07-16-2007

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10:06:14

`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 -->

The derivative of x^2 y, according to the chain rule fg = f' g + g' f, is this:

2x y + y' x^2

confidence assessment: 2

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10:06:23

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 -->

Ok.

self critique assessment: 3

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10:13:10

`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 -->

According to the chain rule in which f( g(x) ) = g' * f'(g(x)), this derivative should be:

y' * 3 x^2

confidence assessment: 2

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10:13:40

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 -->

Ok. I guess I didn't put it all in terms of y.

self critique assessment: 3

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10:29:24

`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 -->

This is basically the last few questions strung together:

y' 3y^2 x^2 + 2x y^3

confidence assessment: 2

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10:36:14

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 -->

Ok.

self critique assessment: 3

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10:38:04

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

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RESPONSE -->

It looks like:

y = (9 - 7x) / 2x^2

confidence assessment: 2

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10:38:38

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 -->

Ok.

self critique assessment: 3

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10:44:21

`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 -->

At x=1:

y = 1

y' = -5.5

confidence assessment: 2

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10:48:52

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 -->

Ok.

self critique assessment: 3

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10:57:25

`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 -->

Continuing the simplification leads to:

4xy + 2x^2 y' + 7 = 0

Solving for y':

y' = (-7 -4xy) / 2x^2 or

y' = -7 / 2x^2 -4y / 2x

When x and y both equal 1:

y' = -7/2 -2 or

y' = -11 / 2

confidence assessment: 2

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10:58:23

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 -->

I did it slightly differently, but my answers were the same in the end.

self critique assessment: 3

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11:27:35

`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 -->

It looks like the derivative equation is:

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

However, this does not equal 0 at x=1 and y=2, and I am not quite sure what to do.

confidence assessment: 1

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13:48:53

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 -->

Hmm. I see where the y' appeared, but I don't understand why. After that I understand the steps.

confidence assessment: 2

For example

d(y^2 x^2) / dx = d(y^2)/dx * x^2 + y^2 * d(x^2)/dx.

y^2 represents the function f(g(x)) = g(x)^2, where g(x) is y(x). The derivative of this function, by the chain rule, is g ' (x) * f ' (g(x)) = g ' (x) * 2 g(x). Since g(x) represents your y(x) function, the derivative is therefore y ' (x) * 2 y(x). We write this as 2 y y'.

In short, we write

d(y^2)/dt = 2 y * dy/dt.

d(x^2)/dx is the derivative of x^2 with respect to x, which could be written 2 x x', where since the derivative is with respect to x, x' = dx /dx. However since the derivative of x with respect to x is just 1, we don't need to write it as part of the expression, and we didn't need the chain rule for this derivative in the first place.

See also the explanation of the preceding problem.

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14:29:29

`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 -->

It looks like it goes like this:

( x^2 sin(y) )' - ( sin( xy ) )' = (0)'

2x sin(y) + (sin(y))' x^2 - ((xy)' cos(xy))

2x sin(y) + x^2 y' cos(y) - ((y'x + y) cos(xy))

2x sin(y) + x^2 y' cos(y) - y cos(xy) - y' x cos(xy)

Isolate y':

y' x^2 cos(y) - y' x cos(xy) = y cos(xy) - 2x sin(y)

y'(x^2 cos(y) - x cos(xy)) = y cos(xy) - 2x sin(y)

y' = ( y cos(xy) - 2x sin(y) ) / ( x^2 cos(y) - x cos(xy) )

At x=3 and y='pi:

y' = ( 'pi cos(3*'pi) - 2(3) sin('pi) ) / ( (3)^2 cos('pi) - (3) cos(3*'pi) )

y' = 0.523598

confidence assessment: 1

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14:53:15

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 -->

Hmm. The sign of my answer is positive, while this answer is negative, and I see why it is negative. My calculator just returned a positive value, I suppose.

self critique assessment: 3

The result should be pi/6.

`pi * -1 - 2 * 3 * 0 = -pi and 9 * -1 - 3 * -1 = - 6, and -pi/(-6) = pi/6.

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