course MTH272 that was a great exercise.really cleared up my understanding of the chain rule.it kind of got me excited for the next exercise because im beginning to understand what i was missing ???????????????assignment #012012. The Chain Rule
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23:33:59 `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 --> y=e^(t^2) z=t^2 z= -4, -1, -.25, 0, .25, 1, 4 y=.018, .368, .779, 1, 1.28, 2.72, 54.60 confidence assessment: 3
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23:34:58 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 wasnt sure on your notation on the first couple numbers. like if the negative should be included or not in the squaring of the number, but the concept is pretty simple self critique assessment: 3
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23:43:29 `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 --> the same as the previous question.its the same formula just composited into one step. 54.6, 2.72 and so on confidence assessment: 3
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23:43:34 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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23:47:52 `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 --> the two simple functions would be y=cos(z) z=ln(x) confidence assessment: 3
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23:48:01 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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23:48:57 `q004. What would be the chain of functions for y = ( ln(t) ) ^ 2?
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RESPONSE --> the simple functions would be y=z^2 z=(ln(t)) confidence assessment: 3
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23:49:02 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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23:49:26 `q005. What would be the chain of functions for y = ln ( cos(x) )?
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RESPONSE --> y=ln(z) z=cos(x) confidence assessment: 3
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23:49:31 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 --> self critique assessment: 3
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23:53:13 `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)) =sin(z) y'= cos(z) z'= 1/x y'=(cos(1/x)) confidence assessment: 3
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23:58:39 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 missed in the previous explanation about moving the g(x) derivative out in front, and leaving the original function inside. lets try that again. g'(x)*f'(g(x) y=sin(ln(x)) y'= cos(ln(x) g'(x)= 1/x y'= 1/x*cos(lnx) self critique assessment: 2
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23:59:58 `q007. Find the derivative of y = ln ( 5 x^7 ) .
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RESPONSE --> y=ln(z) z=5x^7 z'= 35x^6 y'=35x^6*(1/x)(5x^7) confidence assessment: 3
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00:01:37 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 --> im beginning to grasp the chain rule, but in this one i am confused on why you can move the 'z' function into the denominator of the f(x).Shouldnt they multiply?or is that just a simplification? self critique assessment: 2
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00:04:13 `q008. Find the derivative of y = e ^ ( t ^ 2 ).
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RESPONSE --> z= t^2 y=e^z y'=2t*e^(t^2) confidence assessment: 3
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00:04:26 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 --> self critique assessment: 3
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00:05:17 `q009. Find the derivative of y = cos ( e^t ).
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RESPONSE --> y=cos(e^t) z=e^t y'=e^t*-sin(e^t) confidence assessment: 3
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00:05:27 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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00:07:09 `q010. Find the derivative of y = ( ln ( t ) ) ^ 9.
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RESPONSE --> z=lnt y=z^9 z'=1/t y'=1/t * 9(lnt)^8 y'=(9(ln(t))^8)/t confidence assessment: 3
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00:07:17 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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00:08:36 `q011. Find the derivative of y = sin^4 ( x ).
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RESPONSE --> z'=1 y'=cos^4(x) confidence assessment: 2
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00:12:15 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 --> the order confused me on that question lets try again. y=sin^4(x) z=sin(x) y=z^4 y'=4z^3 z'=cos(x) cos(x)*4(sin(x))^3=4(cos(x)sin^3(x) self critique assessment: 2
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00:14:42 `q012. Find the derivative of y = cos ( 3x ).
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RESPONSE --> y=cos(3x) 3x=z cos(z)=y y'=-sin(z) z'=3 y'=3*-sin(3x) y'=-3sin(3x) confidence assessment: 3
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00:15:39 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 --> this exercise really cleared up my understanding of the chain rule.really took me through step by step.it kind of cleared up one of the missing pieces to my calculus self critique assessment: 3
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