qa 12

qa 12

Good job with this assignment. See my notes and be sure you understand all the details.

assignment #012 012. The Chain Rule

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21:17:19

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 --> z = (-2)^2 z=4 y = e^z y=54.6

z = (-1)^2 z=1 y = e^z y=2.7

z = (-.5)^2 z=.25 y = e^z y=1.3

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

z = (.5)^2 z=.25 y = e^z y=1.3

z = (1)^2 z=1 y = e^z y=2.7

z = (2)^2 z=4 y = e^z y=54.6

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21:17:42 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 --> yup.

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21:28:40 `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 --> y = e^((-2)^2) =55

y = e^((-1)^2) =3

y = e^((-.5)^2)

all same as #1, because equation is combined

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21:29:07 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 --> yup again.

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21:32:00 `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 --> t= ln(x)

y = cos ( t )

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21:32:50 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 --> ok.

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21:34:58 `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2?

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RESPONSE --> z = ln(t)

y = z ^ 2

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21:35:38 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 --> yesss

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21:37:12 `q005. What would be the chain of functions for y = ln ( cos(x) )?

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RESPONSE --> z = cos(x)

y = ln(z)

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21:37:31 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 --> yup.

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`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 --> y = sin ( ln ( x ) ) y ' = g' ( x ) * f ' ( g ( x) y' = lnx' * sin'( lnx') y' = 1/x * cos( 1/x) y' = 1/x * cos1/x i think that's it. those damn ghetto puter symbols blend together and take forever to decipher.

It's best to write them out. The notation here is not significantly different than that used in the text. y = f ( g ( x) ) and y ' = g' ( x ) * f ' ( g ( x) ) are exactly as they appear in most textbooks and exactly as they would appear in typical computer-algebra software.

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19:08:44 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 --> crap. so the inner function is not modified?

Right. Good way to put it.

To connect this idea with the definition:

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

g(x) is the same in the expression for y as in the expression for y '. No modification.

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19:18:47 `q007. Find the derivative of y = ln ( 5 x^7 ) .

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RESPONSE --> y = ln ( 5 x^7 )

y = 1/( 5 x^7 )

this is f ' (g(x)).

5 x^7 ' =7(5x)^6

this is g ' (x)

y = 7(5x)^6 * 1/( 5 x^7 )

this is g ' (x) * f ' (g(x) ).

Good job.

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19:20:11 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 --> ok.

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19:24:42 `q008. Find the derivative of y = e ^ ( t ^ 2 ).

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RESPONSE --> y = e ^ ( t ^ 2 )

y' = e ^ ( t ^ 2 )

( t ^ 2 )' = ( 2t ^ 1 ) = 2t

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

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19:25:37 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 --> yup.

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19:28:17 `q009. Find the derivative of y = cos ( e^t ).

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RESPONSE --> y = cos ( e^t )

y' = -sin ( e^t )

( e^t )' = e^t

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

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19:28:33 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 --> ok.

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19:34:31 `q010. Find the derivative of y = ( ln ( t ) ) ^ 9.

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RESPONSE --> y = ( ln ( t ) ) ^ 9

y' =9 [ ln ( t ) ] ^ 8

ln ( t )' = 1/t

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

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19:34:55 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 --> yup.

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19:40:39 `q011. Find the derivative of y = sin^4 ( x ).

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RESPONSE --> y = sin^4 ( x )

y' = cos^4 ( x )

???

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19:44:30 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 --> still doesn't make a lot of sense, but ok.

Once you see that the function is y = f(g(x)) with f(z) = z^4 and g(x) = sin(x) the rest follows.

Make sure you understand how this is the correct 'chain' of functions. If not holler and I'll give you a few more examples.

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19:49:16 `q012. Find the derivative of y = cos ( 3x ).

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RESPONSE --> y = cos ( 3x )

y' = -sin(3x)

3x' = 1(3x^0) ???

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19:51:11 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 --> so the der. of 3x is 3. ok.

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