qa12

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course Mth 271

10/24 9p

012. The Chain Rule

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Question: `qNote that there are 12 questions in this assignment.

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Self-critique (if necessary):

Self-critique Rating:

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Question: `q001. When we form the composite of two functions, we first apply one function to the variable, then we apply the other function to the result.

We can for example first apply the function z = t^2 to the variable t, then we can apply the function y = e^z to our result.

If we apply the functions as specified to the values t = -2, -1, -.5, 0, .5, 1, 2, what y values to we get?

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Your solution:

z=t^2 t t^2 y=e^z z y=

-2 4 4 54.598

-1 1 1 2.718

-.5 .25 .25 1.284

0 0 0 1

.5 .25 .25 1.284

1 1 1 2.718

2 4 4 54.598

confidence rating #$&*:3

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Given Solution:

`aIf t = -2, then z = t^2 = (-2)^2 = 4, so y = e^z = e^4 = 55, approx..

If t = -1, then z = t^2 = (-1 )^2 = 1, so y = e^z = e^1 = 2.72, approx..

If t = -.5, then z = t^2 = (-.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx.

If t = 0, then z = t^2 = ( 0 )^2 = 0, so y = e^z = e^0 = 1.

If t = .5, then z = t^2 = (.5 )^2 = .25, so y = e^z = e^.25 = 1.28, approx.

If t = 1, then z = t^2 = ( 1 )^2 = 1, so y = e^z = e^1 = 2.72, approx..

If t = 2, then z = t^2 = ( 2 )^2 = 4, so y = e^z = e^4 = 55, approx..

Self-critique (if necessary) :

n/a

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Self-critique Rating:3

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Question: `q002. If we evaluate the function y = e^(t^2) at t = -2, -1, -.5, 0, .5, 1, 2, what y values do we get?

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Your solution:

same as #1

confidence rating #$&*:3

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Given Solution:

`aIf t = -2, then y = e^(t^2) = e^(-2)^2 = e^4 = 55, approx.

If t = -1, then y = e^(t^2) = e^(-1)^2 = e^1 = 2.72, approx.

If t = -.5, then y = e^(t^2) = e^(-.5)^2 = e^.25 = 1.28, approx.

If t = 0, then y = e^(t^2) = e^(0)^2 = e^0 = 1.

If t = .5, then y = e^(t^2) = e^(.5)^2 = e^.25 = 1.28, approx.

If t = 1, then y = e^(t^2) = e^(1)^2 = e^1 = 2.72, approx.

If t = 2, then y = e^(t^2) = e^(2)^2 = e^2 = 55, approx.

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Self-critique (if necessary):

n/a

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Self-critique Rating:3

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Question: `q003. We see from the preceding two examples that that the function y = e^(t^2) results from the 'chain' of simple functions z = t^2 and y = e^z.

What would be the 'chain' of simple functions for the function y = cos ( ln(x) )?

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Your solution:

y = cos ln (x)

f g x

u = ln x y = cos (u)

confidence rating #$&*:2

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Given Solution:

`aThe first function encountered by the variable x is the ln(x) function, so we will say that z = ln(x). The result of this calculation is then entered into the cosine function, so we say that y = cos(z).

Thus we have y = cos(z) = cos( ln(x) ).

We also say that the function y(x) is the composite of the functions cosine and natural log functions, i.e., the composite of y = cos(z) and z = ln(x).

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Self-critique (if necessary):I really need more examples of this.

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Self-critique Rating:2

@&

Between these exercises and the text you'll see numerous examples.

In any case your breakdown

y = cos(u)

with

u = ln(x)

is correct.

*@

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Question: `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2?

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Your solution:

y = ln(t) y = t^2 y = (ln(t)^2

confidence rating #$&*:2

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Given Solution:

`aThe first function encountered by the variable t is ln(t), so we say that z = ln(t). This value is then squared so we say that y = z^2.

Thus we have y = z^2 = (ln(t))^2.

We also say that we have here the composite of the squaring function and the natural log function, i.e., the composite of y = z^2 and z = ln(t).

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Self-critique (if necessary):

It is getting clearer.

@&

You pick things up well, so I expect that it will do exactly that.

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Self-critique Rating:2

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Question: `q005. What would be the chain of functions for y = ln ( cos(x) )?

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Your solution:

y = ln(x) y = cos(x) y = ln(cos(x) )

confidence rating #$&*:2

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Given Solution:

`aThe first function encountered by the variable is cos(x), so we say that z = cos(x). We then take the natural log of this function, so we say that y = ln(z).

Thus we have y = ln(z) = ln(cos(x)).

This function is the composite of the natural log and cosine function, i.e., the composite of y = ln(z) and z = cos(x).

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Self-critique (if necessary):

In the book the examples list both functions as y=. Here one is y= and one is z=. Does that make a difference?

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Self-critique Rating:2

@&

Not really, but it's easier to keep track of things if you use z or u for the 'inner' function of the composite.

Using u, this would be

y = ln(u)

with

u = cos(x).

*@

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Question: `q006. The rule for the derivative of a chain of functions is as follows:

The derivative of the function y = f ( g ( x) ) is y ' = g' ( x ) * f ' ( g ( x) ).

For example if y = cos ( x^2 ) then we see that the f function is f(z) = cos(z) and the g function is the x^2 function, so that f ( g ( x) ) = f ( x^2 ) = cos ( x^2 ) . By the chain rule the derivative of this function will be

(cos(x^2)) ' = g ' ( x) * f ' ( g ( x) ) .

g(x) = x^2 so g'(x) = 2 x.

f ( z ) = cos ( z) so f ' ( z ) = - sin( z ), so f ' ( g ( x ) ) = - sin ( g ( x ) ) = - sin ( x^2).

Thus we obtain the derivative

(cos(x^2)) ' = g ' ( x ) * f ' ( g ( x ) ) =

2 x * ( - sin ( x^2 ) ) =

- 2 x sin ( x^2).

Apply the rule to find the derivative of y = sin ( ln ( x ) ) .

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Your solution:

y = sin(x) y= ln(x)

y’ = sin(x) = -cos (x) y’ = ln (x) = 1/x

y’ = sin(ln(x)) = (1/x)(-cos (x)) = -cos(1)

confidence rating #$&*:2

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Given Solution:

`aWe see that y = sin ( ln(x) ) is the composite f(g(x)) of f(z) = sin(z) and g(x) = ln(x). The derivative of this composite is g ' (x) * f ' ( g(x) ).

Since g(x) = ln(x), we have g ' (x) = ( ln(x) ) ' = 1/x.

Since f(z) = sin(z) we have f ' (z) = cos(z).

Thus the derivative of y = sin( ln (x) ) is

y ' = g ' (x) * f ' (g(x)) = 1 / x * cos ( g(x) ) =

1 / x * cos( ln(x) ).

Note how the derivative of the 'inner function' g(x) = ln(x) appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the sine function.

STUDENT QUESTION

We have been given many different derivative functions in the last two assignments are these things that will always be

given or should I make a chart of them and start learning them by memory.

INSTRUCTOR RESPONSE

There are not that many basic functions (power, exponential, logarithmic, sine, cosine cover most of it). You will need to know the derivatives of these functions and maybe a few more as we go along, and the rules for derivatives of product, quotient and composite functions.

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Self-critique (if necessary):

2 things I don’t understand:

Why is the derivative of the cos function -sin but the derivative of the sin function positive?

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@&

You aren't expected to understand the sine and cosine functions in this course. So you just need to accept that the derivatives are as they have been given.

If you do have a background in trigonometry, and especially in the circular definitions of the sine and cosine function, let me know and I'll be glad to explain. If you don't have the background, it won't be possible for me to explain. (Brief statement: It's easy to see from the circular definition that the cosine is decreasing the sine is positive, and when the sine is increasing the cosine is positive.)

The main reason I use those functions in this course is that their form makes it easy to see composites.

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Why do you put the function ln(x) back in instead of the derivative ?

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Self-critique Rating:1

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Question: `q007. Find the derivative of y = ln ( 5 x^7 ) .

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Your solution:

y = ln(x) y = 5x^7

y’ = (1/x) y’ = (7)(5x^6) = 35x^6

y’ = (35x^6)/(x) = 35x^5

confidence rating #$&*:2

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Given Solution:

`aFor this composite we have f(z) = ln(z) and g(x) = 5 x^7. Thus

f ' (z) = 1 / z and g ' (x) = 35 x^6.

We see that f ' (g(x)) = 1 / g(x) = 1 / ( 5 x^7).

So the derivative of y = ln( 5 x^7) is

y ' = g ' (x) * f ' (g(x)) = 35 x^6 * [ 1 / ( 5 x^7 ) ].

Note again how the derivative of the 'inner function' g(x) = 5 x^7 appears 'out in front' of the derivative of the 'outer' function; that derivative is in this case the reciprocal or 1 / z function.

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Self-critique (if necessary):

It looks like I missed substituting the 5x^7 back to the bottom for the x in the denominator.

Wouldn’t this expression simplify to 7/x?

@&

It would. Good observation.

By the laws of logarithms, in fact, ln( 5 x^7) = ln(5) + 7 ln(x).

The derivative of 5 is zero, and the derivative of ln(x) being 1/x we conclude that the derivative of this expression is 7 / x.

*@

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Self-critique Rating:2

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Question: `q008. Find the derivative of y = e ^ ( t ^ 2 ).

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Your solution:

y = e^x y = t^2

y’= e^x y’ = 2t

y’ = e^t^2 = 2t^e

confidence rating #$&*:2

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Given Solution:

`aThis function is the composite of f(z) = e^z and g(t) = t^2. We see right away that f ' (z) = e^z and g ' (t) = 2t.

Thus the derivative of y = e^(t^2) is y ' = g ' (t) * f ' (g(t)) = 2 t * e^(t^2).

Note once more how the derivative of the 'inner function' g(t) = t^2 appears 'out in front' of the derivative of the 'outer' function.

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Self-critique (if necessary):I think I missed the same step as in the last problem. I need a way to remember the last step.

@&

If you carefully use the expression

g ' (x) f ' (g(x))

you'll be more likely to remember to do that substitution.

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Self-critique Rating:2

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Question: `q009. Find the derivative of y = cos ( e^t ).

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Your solution:

y = cos (x) y = e^t

y’ = sin (x) y’ = e^x

y’ = cos(e^t) = sin(e^t)

confidence rating #$&*:2

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Given Solution:

`aWe have the composite of f(z) = cos(z) and z(t) = e^t, with derivatives f ' (z) = -sin(z) and g ' (t) = e^t.

Thus the derivative of y = cos ( e^t ) is y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t).

Note how the 'inner function' is unchanged, as it has been in previous examples, and how its derivative appears in front of the derivative of the 'outer' function.

STUDENT QUESTION: I am confused as to why the end result is - e^t sin (e^t). Is it because we just combined the multiplication?

I believe you are asking why e^t * -sin( e^t) = - e^t sin ( e^t)

e^t * (-sin( e^t)) = (- sin ( e^t) ) * e^t by the commutativity of multiplication

- sin ( e^t) * e^t = - ( sin ( e^t) * e^t) by the laws for multiplying signed numbers.

- ( sin ( e^t) * e^t) = - ( e^t * sin(e^t) ) by commutativity

- ( e^t * sin(e^t) ) = - e^t * sin(e^t) by the laws for multiplying signed numbers

So e^t * -sin( e^t) = - e^t sin ( e^t).

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Self-critique (if necessary):

I see why I have been missing the f’ of cos(z) = -sin(z). I went back to the notes I took in qa11 and then I went back to qa11 and I did miss the negative when I took the notes.

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Self-critique Rating:

@&

Good.

Except for these exercises you won't be taking derivatives of expressions involving sines and cosines. They are used here because they nicely illustrate the rules of differentiation.

*@

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Question: `q010. Find the derivative of y = ( ln ( t ) ) ^ 9.

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Your solution:

y’ =(ln(t)) = 1/t y’ = (ln(t))^9 = 9(ln(t))^8

y’= (ln(t))^9 = (1/t)[9(ln(t))]^8

confidence rating #$&*:

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Given Solution:

`aWe have y = f(g(t)) with f(z) = z^9 and g(t) = ln(t). f ' (z) = 9 z^8 and g ' (t) = 1 / t. Thus

y ' = g ' (t) * f ' (g(t)) = 1/t * 9 ( ln(t) )^8.

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Self-critique (if necessary):n/a

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Self-critique Rating:3

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Question: `q011. Find the derivative of y = sin^4 ( x ).

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Your solution:

y’ = sin^(4)(x) = cos^4(x) = 4 cos^3(x)

confidence rating #$&*::2

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Given Solution:

`aThe composite here is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x). Note that the notation sin^4 means to raise the value of the sine function to the fourth power.

We see that f ' (z) = 4 z^3 and g ' (x) = cos(x). Thus

y ' = g ' (x) * f ' (g(x)) = cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x).

STUDENT QUESTION: I am having a hard time figuring out the simplication of this one? Any help?

INSTRUCTOR RESPONSE: You had the right function for the f and g functions, except that you reversed them:

If f(z) = sin(z) and g(x) = x^4, then f(g(x)) = sin(x^4), and the derivative would indeed be 4 x^3 * cos(x^4).

However the function here is sin^4(x), so f(z) = z^4 and g(x) = sin(x). This gives a very different result, as shown in the given solution.

Regarding the simplification step cos(x) * 4 ( sin(x) ) ^ 3 = 4 cos(x) sin^3 (x):

a * b * c = b * a * c; this actually requires multiple applications of associative and commutative laws for addition, but it's something we're used to so we don't usually think about it that deeply. We know it is 'safe' to change the order of a sequence of multiplications.

cos(x) * 4 ( sin(x) ) ^ 3 is just a sequence of multiplications. If a = cos(x), b = 4 and c = (sin(x))^3, then this expression is a * b * c.

Then b * a * c = 4 cos(x) sin^3 (x).

Note that sin^3(x) means (sin(x))^3.

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Self-critique (if necessary):

I don’t understand where the f(z) notation comes from. It’s not in the original problem, so how do you know to add a f(z) to break the functions down?

Also, I don’t understand why the sin has to stay in there.

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self-critique Rating:1 f g

@&

It can be confusing to use x for two different purposes. So we introduce a 'dummy' variable like z or u to avoid errors and clarify what we're doing. The 'dummy' variable doesn't appear in the final expression; it's used only to facilitate the process and disappears when we do our substitutions.

*@

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Question: `q012. Find the derivative of y = cos ( 3x ).

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Your solution:

f(x) = cos x f’(x) = -sin(x)

g(x) = 3x g’(x) = 3

g’(x) * f’(g(x)) = 3 - sin(3x)

confidence rating #$&*:2

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Given Solution:

`aThis is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus

y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x).

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Self-critique (if necessary):

Still feel like I need much more practice. I noticed that the chain rule is not covered in the notes and book until much later.

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Self-critique Rating:2

@&

It isn't covered for awhile, but having covered it here, where it is certainly possible to do so, it will be much less daunting when you encounter it in the text.

*@

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Question: `q012. Find the derivative of y = cos ( 3x ).

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Your solution:

f(x) = cos x f’(x) = -sin(x)

g(x) = 3x g’(x) = 3

g’(x) * f’(g(x)) = 3 - sin(3x)

confidence rating #$&*:2

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Given Solution:

`aThis is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus

y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x).

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Self-critique (if necessary):

Still feel like I need much more practice. I noticed that the chain rule is not covered in the notes and book until much later.

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Self-critique Rating:2

@&

It isn't covered for awhile, but having covered it here, where it is certainly possible to do so, it will be much less daunting when you encounter it in the text.

*@

#*&!

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Question: `q012. Find the derivative of y = cos ( 3x ).

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Your solution:

f(x) = cos x f’(x) = -sin(x)

g(x) = 3x g’(x) = 3

g’(x) * f’(g(x)) = 3 - sin(3x)

confidence rating #$&*:2

^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^

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Given Solution:

`aThis is the composite y = f(g(x)) with f(z) = cos(z) and g(x) = 3x. We obtain f ' (z) = - sin(z) and g ' (x) = 3. Thus

y ' = g ' (x) * f ' (g(x)) = 3 * (-sin (3x) ) = - 3 sin(3x).

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Self-critique (if necessary):

Still feel like I need much more practice. I noticed that the chain rule is not covered in the notes and book until much later.

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Self-critique Rating:2

@&

It isn't covered for awhile, but having covered it here, where it is certainly possible to do so, it will be much less daunting when you encounter it in the text.

*@

#*&!#*&!

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