Q_A 12

course Mth 271

???????z???Student Name:

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assignment #012

012. The Chain Rule

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14:40:23

Note 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 -->

-2^2=4

y= e^4

=54.6

-1^2= 1

y=e^1

= 2.7

-.5^2=.25

y=e^.25

= 1.3

0^2=0

y= e^0

=1

.5^2= .25

y= 1.3

y= 2.7

y= 54.6

They start to go over and over again

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14:40:46

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 -->

I understand

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14:42:22

`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 -->

t= -2

-2^2= e^4= 55

t= -1

-1^2= 2.7

You get the exact same thing over and over again like the first thing we did...

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14:43:05

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 -->

This is what I got but I just didnt type it all out again when I realized that I was getting the same thing as the first one......but I understand this

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14:44:15

`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 -->

y= cos(z)

z=In(x)

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14:44:34

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 -->

I understand this very well

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14:45:28

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

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RESPONSE -->

y= z^2

z= (In(t))

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14:45:46

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 -->

ok

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14:46:20

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

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RESPONSE -->

y= In(z)

z= cos(x)

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14:46:41

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 -->

got it

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14:49:26

`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 -->

OK......I am having an issue with this......I cant understand the f(x) and the g(x).....it's got me messed up........I think that it wouls be

1/x* cos (In(x))

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14:50:43

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 -->

I understand this to some degree but I am having a problem with what you are trying to tell me with the g and f......could I not say take the ' of the middle and then take the ' of the thing as a whole

I know of no more succinct and consistent way to express this. In order to express the pattern, we need a general notation, which is consistent from problem to problem, to refer to 'this function' and 'that function', or more specifically to 'the outer function f(x) and the inner function g(x)'.

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14:51:48

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

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RESPONSE -->

35x^6 * 1( 1/ 5*x^7)

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14:52:10

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 -->

I guess that it is the end and then the whole thing

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14:53:19

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

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RESPONSE -->

2t * e^ (t^2)

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14:53:26

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 -->

got it

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14:54:14

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

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RESPONSE -->

e^t *-sin(e^t)

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14:54:43

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 -->

I should have put the - sign out it infront but I left it with the sin

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14:55:43

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

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RESPONSE -->

9 *1/t* (In(t)) ^8

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14:55:50

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 -->

got that

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14:56:37

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

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RESPONSE -->

4 sin^3 (x)

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14:57:48

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 -->

I knew that there had toi be more to it I just got lost with it.....I dont know why there should be a sin and a cos in the problem I understand that I should have had a cos but not sure about the sin

The derivative of the sin function is the cosine function. If g(x) = sin(x) and f(z) = z^4, then g'(x) is cos(x) and f ' (z) = 4 z^3, so g '(x) * f ' (g(x)) = 4 cos(x) sin^3(x).

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14:58:18

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

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RESPONSE -->

3 -sin(3x)

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14:58:45

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 -->

I should have known to put the - out in front this time but I just forgot

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3 -sin(3x) isn't the same a -3 sin(3x).

3 * (-sin(3x)) would be the same thing, though -3 sin(3x) is still the standard simplification.

Good, but see my notes. Let me know if you have questions.