first 12

course Mth 173

p??h???x?????assignment #012012. The Chain Rule

Your work has been received. Please scroll through the document to see any inserted notes (inserted at the appropriate place in the document, in boldface) and a note at the end. The note at the end of the file will confirm that the file has been reviewed; be sure to read that note. If there is no note at the end, notify the instructor through the Submit Work form, and include the date of the posting to your access page.

02-23-2007

......!!!!!!!!...................................

09:36:49

`qNote that there are 12 questions in this assignment.

`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?

......!!!!!!!!...................................

RESPONSE -->

z = t ^ 2

z = -2 ^ 2

z = 4

y = e ^ 4

y = 54.6

z = t ^ 2

z = -1 ^ 2

z = 1

y = e ^ 1

y = 2.72

z = t ^ 2

z = -.5 ^ 2

z = .25

y = e ^ .25

y = 1.28

z = t ^ 2

z = 0 ^ 2

z = 0

y = e ^ 0

y = 1

z = t ^ 2

z = -1 ^ 2

z = 1

y = e ^ 1

y = 2.72

z = t ^ 2

z = .5 ^ 2

z = .25

y = e ^ .25

y = 1.28

z = t ^ 2

z = 1 ^ 2

z = 1

y = e ^ 1

y = 2.72

z = t ^ 2

z = 2 ^ 2

z = 4

y = e ^ 4

y = 54.6

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:36:54

If 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..

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:39:07

`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?

......!!!!!!!!...................................

RESPONSE -->

y = e^(t^2)

y = e^(-2^2)

y = e^(4)

y = 54.6

y = e^(t^2)

y = e^(-1^2)

y = e^(1)

y = 2.72

y = e^(t^2)

y = e^(-.5^2)

y = e^(.25)

y = 1.28

y = e^(t^2)

y = e^(0^2)

y = e^(0)

y = 1

y = e^(t^2)

y = e^(.5^2)

y = e^(.25)

y = 1.28

y = e^(t^2)

y = e^(1^2)

y = e^(1)

y = 2.72

y = e^(t^2)

y = e^(2^2)

y = e^(4)

y = 54.6

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:39:11

If 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.

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:41:08

`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) )?

......!!!!!!!!...................................

RESPONSE -->

z = ln(x)

y = cos(z)

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:41:14

The 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).

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:42:46

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

......!!!!!!!!...................................

RESPONSE -->

z = ln(t)

y = z ^ 2

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:42:50

The 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).

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:44:09

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

......!!!!!!!!...................................

RESPONSE -->

z = cos(x)

y = ln(z)

confidence assessment: 3

.................................................

......!!!!!!!!...................................

09:44:13

The 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).

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:48:48

`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 ) ) .

......!!!!!!!!...................................

RESPONSE -->

y = sin (ln(x))

f(z) = sin(z)

g(x) = ln(x)

y' = g'(x) * f'(g(x))

y' = 1/x * cos(ln(x))

confidence assessment: 2

.................................................

......!!!!!!!!...................................

09:48:52

We 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.

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:51:25

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

......!!!!!!!!...................................

RESPONSE -->

y = ln(5 x^7)

f(z) = ln(z)

g(x) = 5 x ^ 7

y' = f'(g(x))

y' = 1 / 5 x ^ 7

y ' = 35 x ^ 6 (1 / 5 x ^ 7)

1 / 5 x ^ 7 means 1/5 * x^7, by the order of operations.

Some authors use 'implied parentheses', but this leads to ambiguity and potential confusion.

If the expression 1 / 5 x^7 is put into a computer algebra system it will be displayed as x^7 / 5, with a strict interpretation of the order of operations.

To avoid confusion you should write the denominator 5 x^7 grouped, as shown below.

confidence assessment: 2

.................................................

......!!!!!!!!...................................

09:51:29

For 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.

The expression given above can of course be simplified to 7 / x; the simplified expression was not given in this solution because I wanted to emphasize the structure of the process.

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

09:52:51

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

......!!!!!!!!...................................

RESPONSE -->

y = e ^ (t ^ 2)

f(z) = e ^ z

g(x) = t ^ 2

y' = 2t * (e^(t^2))

confidence assessment: 2

.................................................

......!!!!!!!!...................................

09:52:55

This 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.

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:01:36

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

......!!!!!!!!...................................

RESPONSE -->

y = cos (e^ t)

f(z) = cos(z)

g(x) = e ^ t

y' = e ^ t * (-sin (e ^ t))

y' = -e^t * (sin(e^t))

confidence assessment: 1

.................................................

......!!!!!!!!...................................

10:01:40

We 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.

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:03:31

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

......!!!!!!!!...................................

RESPONSE -->

y = (ln(t)) ^ 9

f(z) = z ^ 9

g(x) = ln(t)

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

confidence assessment: 1

.................................................

......!!!!!!!!...................................

10:03:34

We 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.

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

You left out the intermediate steps

f ' (z) = 9 z^8 and

g ' (t) = 1 / t.

If you can think through them, as apparently you can, then they aren't absolutely necessary. With practice, you can think through these steps reliably. But I don't recommend skipping them at this point.

.................................................

......!!!!!!!!...................................

10:05:31

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

......!!!!!!!!...................................

RESPONSE -->

y = sin ^ 4 (x)

f(z) = z ^ 4

g(x) = sin(z)

y' = cos(z) * 4(sin(z))^3

y' = 4 cos(z) * sin^3(z)

confidence assessment: 2

.................................................

......!!!!!!!!...................................

10:05:34

The 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).

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

......!!!!!!!!...................................

10:06:28

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

......!!!!!!!!...................................

RESPONSE -->

y = cos(3x)

f(z) = cos(z)

g(x) = 3x

y' = 3(-sin(3x))

y' = -3 sin(3x)

confidence assessment: 2

.................................................

......!!!!!!!!...................................

10:06:31

This 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).

......!!!!!!!!...................................

RESPONSE -->

self critique assessment: 3

.................................................

"

Looks great. You left out intermediate steps in several of these problems, so be sure you see my cautionary note; but I'm certainly not going to argue with success.