qa 16

course mth 271

mr smith: the woman who was tutoring me is no longer tutoring. i just dont understand the chain rule, and i can't look at a problem and break it up into f(x) and g(x) nor can i remember which rules to use in which situations. so i dont know what to do. shelley

I can 'talk you through' this with a series of questions. It will probably take two or three exchanges to get it. Here's the first set. Send me a copy of the following questions with your answers inserted:

First I want to see if you understand function notation:

If f(z) = z^2, then what are the values of f(2), f(7), f(a + h), f(q - p), f(aardvark) and f(g(x))?

Next I want you to think through, step by step, what you would do to evaluate a series of expressions:

If you are given the value of x, then if you want to find the value of sqrt(x) * e^x, what is the first thing you would do? What is the next thing you would do? What further steps, if any are necessary, would you take to get the value of the expression?

If you are given the value of x, then if you want to find the value of (x - 3)^2, what is the first thing you do with that value of x? What is the next thing? What further steps, if any are necessary, would you take to get the value of the expression?

If you are given the value of x, then if you want to find the value of (e^x) / (sqrt(x)), what is the first thing you do with that value of x? What is the next thing? What further steps, if any are necessary, would you take to get the value of the expression?

In the above, in one case you evaluated two different functions for x then multiplied the results, in another you evaluated two different functions for x and divided the result, and in another you evaluated one function for x and then plugged the resulting value into another function to get your final result. Which was which? In each case there were two functions involved. What were the two functions for each?

Each of the following involves either evaluating two different functions for x then multiplying the results, evaluating two different functions for x and dividing the result, or evaluating one function for x and then plugging the resulting value into another function to get your final result. For each, identify the two functions involved and state what you did with those two functions:

sqrt(x) * ln(x)
e^(x^2)
(x^2-5) / ln(x)
sqrt(e^x)
(x^3 - 4 x + 1) * e^x
e^(sqrt(x)
x^2 / e^x

11-08-2007

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09:07:43

`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 -->

if y'=x, then the deriv of x^2y

is 2x(x)=2x2

confidence assessment: 2

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09:09:31

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 -->

self critique assessment: 3

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09:12:16

`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 -->

y'=3y^2

confidence assessment: 1

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09:17:30

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 -->

some of this is over my head

self critique assessment: 0

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09:18:52

`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 -->

2x(y^3) + x^2(y'(3y^2))

confidence assessment: 0

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09:19:48

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 -->

self critique assessment: 3

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09:20:54

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

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RESPONSE -->

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

confidence assessment: 3

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09:21:11

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 -->

yup

self critique assessment: 3

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09:23:13

`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 -->

i dont know

if you plug x in for one then y'=0 b/c all you have are constants

confidence assessment: 0

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09:27:06

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 -->

what rule is that, i cant find it

self critique assessment: 0

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09:29:09

`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 -->

i cant follow the math

confidence assessment: 0

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09:31:29

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 dont see how you go from 2x^2y+7x=9

to 2(2xy+x^2y')+7=0

i get that the deriv of 7x=7 and the deriv of 9 is zero

self critique assessment: 0

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09:32:11

`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 -->

i cant do this problem b/c i dont get the preceding one

confidence assessment: 0

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09:32:35

`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

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RESPONSE -->

????

confidence assessment: 0

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09:32:48

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

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RESPONSE -->

confidence assessment: 0

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09:32:57

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

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RESPONSE -->

confidence assessment: 0

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09:33:04

y ' ( 6 x^2 y^2 - 6 x y) = -4 x y^3 + 3 y^2. Dividing both sides by the coefficient of y ':

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RESPONSE -->

confidence assessment: 0

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09:33:12

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

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RESPONSE -->

confidence assessment: 0

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09:33:28

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 -->

confidence assessment: 0

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09:33:43

`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 -->

skip

confidence assessment: 0

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09:33:55

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 -->

n/a

self critique assessment: 0

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You aren't getting this and will need to repeat this Query, but first let's get the rules down. See the questions I posed at the beginning of the document and respond as indicated. It might take two or three sets of questions, but you'll get it.