course mth173 ??x??J?w?????????assignment #012012. The Chain Rule
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01:06:25 `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 --> The formula of z=t^2 and y=e^z.. Plugging in the points for x we get; t z y -2 -4 .018 -1 -1 .368 -.5 -.25 .779 0 0 1 .5 .25 1.28 1 1 2.78 2 4 54.60 confidence assessment: 2
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01:11:34 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 --> my numbers for the negative t values where not correct because as I squared them I didn't enter the ""-"" in the parenthesis, I should have caught the mistake when I wrote down several times that -2^2 = -4 but I didn't. The -t values are the same as the corresponding +t values this would indicate a parabola we were to graph it. self critique assessment: 2
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01:15:56 `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 --> we get the same y values as before. t y -2 54.6 -1 2.72 -.5 1.28 0 1 .5 1.28 1 2.72 2 54.6 confidence assessment: 3
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01:17:14 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 --> There doesn't appear to be a difference between combining the two equations and completing one equation and then plugging that answer into the other equation self critique assessment: 3
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01:19:29 `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) When combining the equation parenthesis are used to break up the different simple equations and get them done in the order that they need to be done. confidence assessment: 3
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01:20:23 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 --> okay self critique assessment: 3
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01:21:39 `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2?
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RESPONSE --> z = ln(t) and z^2 = y again seperated by the parenthesis. confidence assessment: 3
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01:22:17 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 --> okay self critique assessment: 3
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01:24:16 `q005. What would be the chain of functions for y = ln ( cos(x) )?
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RESPONSE --> The x variable encounters the cosine function first so z= cos (x) then y= ln(z). We could also say the natural log of the cosine function of x. confidence assessment: 3
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01:24:38 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 --> okay self critique assessment: 3
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01:37:20 `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)) then the function can be broken down into: f(g(x))= sin (ln(x)) g(x)= ln(x) y'= g ' ( x ) * f ' ( g ( x ) ) y'= (1/x) (cos(ln(x))) y'= (cos(ln(x)) / (x) confidence assessment: 2
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01:39:19 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 --> okay self critique assessment: 3
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01:44:19 `q007. Find the derivative of y = ln ( 5 x^7 ) .
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RESPONSE --> y= ln(5x^7) f(g(x)) = ln (5x^7) g(x) = 5x^7 f(g(x))'= 1/(5x^7) g(x)'= 35x^6 y'= g ' ( x ) * f ' ( g ( x ) ) y'= (35x^6) * (1/5x^7) confidence assessment: 2
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01:45:44 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 notice how the g(x) function is out front. self critique assessment: 3
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01:49:47 `q008. Find the derivative of y = e ^ ( t ^ 2 ).
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RESPONSE --> y = e ^ ( t ^ 2 ) f(g(x)) = e ^ ( t ^ 2 ) g(x) = t ^ 2 f(g(x))' = e ^ ( t ^ 2 ) g(x)' = 2t y'= g ' ( x ) * f ' ( g ( x ) ) y'= (2t) * (e^(t^2)) confidence assessment: 2
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01:50:18 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 --> okay self critique assessment: 3
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01:59:15 `q009. Find the derivative of y = cos ( e^t ).
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RESPONSE --> y = cos ( e^t ) f(g(x)) = cos ( e^t ) g(x) = e^t f(g(x))' = -sin ( e^t ) g(x)' = e^t y=' (g(x)) * (f(g(x)) y=' (e^t) (-sin ( e^t )) confidence assessment: 2
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02:00:51 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 --> okay self critique assessment: 3
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02:06:29 `q010. Find the derivative of y = ( ln ( t ) ) ^ 9.
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RESPONSE --> y = ( ln ( t ) ) ^ 9 f(gx)) = ( ln ( t ) ) ^ 9 g(x)) = ln ( t ) f(gx))' = 9( ln ( t ) ) ^ 8 g(x))' =1/ t y'= g(x)' * (f(g(x))' y'= (1/ t) * ( 9( ln ( t ) ) ^ 8) confidence assessment: 2
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02:06:45 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 --> okay self critique assessment: 3
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02:11:25 `q011. Find the derivative of y = sin^4 ( x ).
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RESPONSE --> I think that you meant to put sin(x)^4 so I'm going to derive from that. f(g(x)) = sin(x)^4 g(x) = sin(x) f(g(x))' = 4sin(x)^3 g(x)' = cos(x) y'= (cos(x)) * (4sin(x)^3) confidence assessment: 2
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02:16:45 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 didn't understand the raising the sine to the fourth power I thought that it was a typo but it wasn't. But even though the derivative worked out to be the same. self critique assessment: 2
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02:20:40 `q012. Find the derivative of y = cos ( 3x ).
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RESPONSE --> y = cos ( 3x ) f(g(x)) = cos ( 3x ) g(x) = 3x f(g(x))' = -sin ( 3x ) g(x)' = 3 y'= (3) * (-sin(3x)) y'= -3sin(3x) confidence assessment: 2
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02:20:58 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 --> okay self critique assessment: 3
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