Assignment 12 QA

course Mth 271

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09:29:14

`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?

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RESPONSE -->

t = -2, z = t^2 = (-2)^2 = 4, y = e^z = e^4 = 54.598

t = -1, z = t^2 = (-1 )^2 = 1, y = e^z = e^1 = 2.718

t = -.5, z = t^2 = (-.5 )^2 = .25, y = e^z = e^.25 = 1.284

t = 0, z = t^2 = ( 0 )^2 = 0, y = e^z = e^0 = 1

t = .5, z = t^2 = (.5 )^2 = .25, y = e^z = e^.25 = 1.284

t = 1, z = t^2 = ( 1 )^2 = 1, y = e^z = e^1 = 2.718

t = 2, z = t^2 = ( 2 )^2 = 4, y = e^z = e^4 = 54.598

confidence assessment: 2

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09:29:19

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..

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RESPONSE -->

self critique assessment: 2

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09:29:36

`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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RESPONSE -->

e^(-2)^2 = e^4 = 54.598

e^(-1)^2 = e^1 = 2.718

e^(-.5)^2 = e^.25 = 1.284

e^(0)^2 = e^0 = 1

e^(.5)^2 = e^.25 = 1.284

e^(1)^2 = e^1 = 2.718

e^(2)^2 = e^2 = 54.598

confidence assessment: 3

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09:29:42

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.

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RESPONSE -->

self critique assessment: 3

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09:29:54

`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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RESPONSE -->

z = ln(x)

y = cos(z)

confidence assessment: 3

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09:29:58

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).

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RESPONSE -->

self critique assessment: 3

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09:30:08

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

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RESPONSE -->

z = ln(t)

y = z^2

confidence assessment: 3

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09:30:10

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).

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RESPONSE -->

self critique assessment: 3

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09:30:21

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

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RESPONSE -->

z = cos(x)

y = ln(z)

confidence assessment: 3

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09:30:24

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).

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RESPONSE -->

self critique assessment: 3

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09:30:41

`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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RESPONSE -->

f(z) = sin(z)

g(x) = ln(x)

g ' (x) = ( ln(x) ) ' = 1/x

f ' (z) = cos(z)

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

confidence assessment: 2

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09:30:45

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.

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RESPONSE -->

self critique assessment: 2

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09:31:00

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

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RESPONSE -->

f(z) = ln(z)

g(x) = 5 x^7

f ' (z) = 1 / z

g ' (x) = 35 x^6

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

confidence assessment: 2

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09:31:03

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.

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RESPONSE -->

self critique assessment: 2

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09:31:44

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

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RESPONSE -->

f(z) = e^z

g(t) = t^2

f ' (z) = e^z

g ' (t) = 2t

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

confidence assessment: 3

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09:31:49

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.

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RESPONSE -->

self critique assessment: 3

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09:32:01

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

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RESPONSE -->

f(z) = cos(z)

z(t) = e^t

f ' (z) = -sin(z)

g ' (t) = e^t

y ' = g ' (t) * f ' (g(t)) = e^t * -sin( e^t) = - e^t sin ( e^t) 3

confidence assessment:

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09:32:04

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.

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RESPONSE -->

self critique assessment: 3

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09:32:15

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

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RESPONSE -->

f(z) = z^9

g(t) = ln(t)

f ' (z) = 9 z^8

g ' (t) = 1 / t

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

confidence assessment: 3

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09:32:18

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.

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RESPONSE -->

self critique assessment: 3

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09:32:28

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

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RESPONSE -->

f(z) = z^4

g(x) = sin(x)

f ' (z) = 4 z^3

g ' (x) = cos(x)

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

confidence assessment: 3

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09:32:30

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).

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RESPONSE -->

self critique assessment: 3

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09:32:40

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

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RESPONSE -->

f(z) = cos(z)

g(x) = 3x

f ' (z) = - sin(z)

g ' (x) = 3

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

confidence assessment: 3

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09:32:42

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).

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RESPONSE -->

self critique assessment: 3

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