course mth 173 assignment #020?????D????????Calculus I
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06:09:43 Query problem 3.6.12 (3d edition 3.6.44). was 4.6.12 derivative of e^( ln(x) + 1) What is the derivative of the given function?
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RESPONSE --> Using the chain rule we get: f(g(x))= e^(ln(x)+1) f(z)= e^z z=g(x)= ln(x)+1 f(g(x))'= f'(g(x)) * (g(x)' = e^(ln(x) +1) * (1/x) = e^(ln(x) +1) / x
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06:12:00 Explain what rule or rules you used to obtain your derivative and how you used them.
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RESPONSE --> I used the chain rule to sperate the function into it's seperate functions, and I used the known derivatives of the functions then used the chain rule to put them back into one function.
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06:13:55 ** This is a composite with f(z) = e^z and g(x) = ln(x) + 1. g'(x) = 1/x, f'(z) = e^z. So the derivative is (e^(ln(x)+1)) ' = (f(g(x)) ' = g'(x) * f'(g(x)) = 1/x e^(ln(x)+1)) = 1/x e^(ln(x)) * e^1 = 1/x * x * e = e. **
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RESPONSE --> I could and should have simplified the problem further knowing that the ln(x) and e^(x) are inverse functions.
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06:16:17 Why is in easier to calculate the derivative of this function if you simplify the function first?
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RESPONSE --> Because using the chain rule you can find the derivative of two ""simple"" function and put the two siimple derivatives together instead of finding the derivative of 1 ""complex"" function
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06:20:19 ** e^(ln x + 1) = e^(ln x) * e^1 = x * e or just e * x. e is a constant so the derivative of e * x is e * 1 = e. **
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RESPONSE --> okay
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06:24:45 Query problem 3.6.43 (3d edition 3.6.44) was 4.6.30 y = ln(x) at x = 1 What is the equation of the tangent line at the given point?
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RESPONSE --> y= ln(x) y'= (1/x) x y y' 1 0 1 given that we have point (1,0) with a slope at that point of 1 we can get the tangent line equation. Ytan=mx +b 0 = 1(1) + b -1=b Ytan = 1(x) - 1
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06:27:01 ** tan line at x = 1 passes through the corresponding graph point (1, ln(1) ) = (1, 0). Slope of tan line equals derivative: y' = 1/x; at x = 1 we have y' = 1. So line has slope 1 and passes through (1, 0). The equation of the line is y - 0 = 1 * (x - 1), or y = x-1. **
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RESPONSE --> again my tangent line equation could have been simplified further, but overall it is the same equation.
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06:30:03 What are your approximations to ln(1.1) and ln(2), based on the tangent line?
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RESPONSE --> Ytan = x -1 x Ytan 1.1 .1 2 1
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06:32:41 ** to get approx ln(1.1): y = 1.1 - 1 = .1. to get approx ln(1.2): y = 1.2 - 1 = .2. Actual values are ln(1.1)=.095 and ln(1.2)=.18. Note that both are a little below the approximations given by the tangent line. **
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RESPONSE --> Sorry I thought it said ln(2) instead of ln(1.2) I thought that it was a little far away for the tangent approximation. Yes the actual values are a little lower because the graph turns away from the tangent line as the x leaves the tangent point.
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06:37:02 In terms of the graph, explain whether your approximations are larger or smaller than the true values.
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RESPONSE --> The tangent line is at point x=1 when x =1 the tangent line is exact because it only touches the graph at that point, the graph moves away from the tangent line as x leaves 1, so the values of the actual function are slightly lower near x=1 and progressivly get ""more"" lower as we move away.
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06:37:18 The value at y=x-1 is higher than the value at y=lnx because the slope of the tangent line works well for values very close to the original point, but there is a limit to its accuracy after moving away so far from the point
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RESPONSE --> okay
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06:37:30 ** Since the graph of y = ln(x) is always concave downward, it will always fall below its tangent-line approximation. **
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RESPONSE --> okay
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06:37:47 ** the graph of the natural log function is concave downward, so the tan line must live above the graph of the actual function. Thus the approximation given by the tangent line will always be high. **
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RESPONSE --> okay
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06:57:28 Query problem 3.7.16 (was 3.7.9 was 4.7.6) e^(x^2) + ln y = 0
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RESPONSE --> e^(x^2) + ln(y) = 0 We will use the chain rule to find the first part of the function. f(g(x))= e^(x^2) f(z) = e^z g(x)= x^2 f(g(x))'= (e^(x^2)) * (2x) we can add this to the larger function (e^(x^2) * 2x) + (ln(y))' = 0 (ln(y))' = -(e^(x^2) * 2x) y' = e^(-(e^(x^2) * 2x)
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07:02:05 When differentiating with respect to x any x terms are differentiated as usual, to differentiate y with respect to x assumes that y is a function of x, in which case its derivative with respect to x is the unspecified quantity dy/dx. The derivative of a function of y is therefore the derivative of a composite. For example cos(y) is the composite of f(y) = cos(y) with the function g(x) = y(x), whose form is not specified. Do g ' (x) * f ' (g(x) ) is y ' (x) f ' (y ( x) ); in the present example this would be y ' (x) * ( -sin(y(x)), which we would write -sin(y) y ' with the understanding that y stands for y(x) and y ' for y ' (x). When we use implicit differentiation to solve for y ' we will typically get some terms containing y ' as a factor and others which don't. We separate these terms algebraically, with the y ' terms on one side and everything else on the other. We then factor out y ' and divide both sides by the other factor to obtain an expression for y ' in terms of y and x. MORE SPECIFIC SOLUTION: The derivative of the equation with respect to x is 2x e^(x^2) + 1/y * y ' = 0. Solving this for y ' we get 1/y * y ' = 2 x e^(x^2) so that y ' = 2x e^(x^2) * y. To see why the derivative of ln y is y ' * 1/y: y is itself a function of x, so ln(y) means ln(y(x)). y(x) is the inner function and its derivative is y'(x) = dy/dx. f(z) = ln(z) is the outer function and its derivative is 1/z. Thus the derivative of f(y(x)) is y'(x) f'(y(x)) = dy/dx * 1/y(x), written just dy/dx * 1/y or y' * 1/y.
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RESPONSE --> so when finding the derivative of a value (y) then you can simply find the normal derivative as if it were x and add y' to it. ex.( ln(y)= (ln(y))' = 1/y * y' or y^2 = (y^2)'= 2y * y' Is this pattern that I'm noticing correct or is it just a coincedence that the situaions that I have seen have played out this way?
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07:10:27 Query problem 3.7.34 (3d edition 3.7.26) was 4.7.18 circle x^2+y^2=25 What are the equations of the tangent lines to the circle at the points where x = 4?
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RESPONSE --> x^2 + y^2 = 25 2x + 2y * y' = 0 2y * y'= -2x y' = (-2x) / (2y) x y y' 4 3 -12 3 = (-1.3) (4) +b 3 = -5.2 +b 8.2 = b Ytan = (-12)x + 8.2
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07:14:15 ** Solving x^2+y^2=25 for x = 4 you get y^2 = 9, which has solutions y = +3 and y = -3. So the points are (4, 3) and (4, -3). By the geometry of the circle the tangent line at a point on the circle is perpendicular to the radial line to that point. The radial lines go from (0,0) to (4,-3) and to (4,3) and so have slopes -3/4 and 3/4, respectively. The tangent lines, being perpendicular, will have the negative reciprocal slopes of 4/3 and -4/3, respectively. Alternatively implicit differentiation of the equation gives us 2 x + 2 y dy/dx = 0, so that dy/dx = -x/y. At (4,3) and (4,-3) we get dy/dx = -4/3 and +4/3, respectively, confirming the previous results. Thus, either way, slope at (4,3) is -4/3, slope at (4,-3) is +4/3. Equation of tangent line at (4,3) is y - 3 = -4/3 ( x - 4) so y = -4/3 x + 25/3. Equation of tangent line at (4,-3) is y - -3 = 4/3 ( x - 4) so y = 4/3 x - 25/3. When x = 4 we have 4^2 + y^2 = 25 so y^2 = 25 - 16 = 9 and y = 3 or -3. Thus the points where x = 4 are (4,3) and (4,-3). The slope from the center (0,0) to (4,3) is 4/3; the tangent line is perpendicular to the radial line from the center to (4,3), so has slope -4/3. The equation of the tangent line is therefore y - 3 = -4/3 (x-4), or y = -4/3 x + 25/3. The slope from the center (0,0) to (4,-3) is -4/3; the tangent line is perpendicular to the radial line from the center to (4,-3), so has slope 4/3. The equation of the tangent line is therefore y - (-3) = 4/3 (x-4) or y = 4/3 x - 25/3. The normal lines are perpendicular to the tangent lines. They therefore have slopes which are the negative reciprocals of the slopes of the tangent lines. Thus the normal lines pass through (4,3) with slope 3/4 and through (-4/3) with slope -3/4. You might note that lines through the origin and with the specified slopes pass through the corresponding points, so those lines are y = 3/4 x and y = -3/4 x, respectively. If you don't notice this you will go ahead and use the point-slope form of the equations. You get y - 3 = 3/4 ( x - 4), which on solving for y gives you y = 3/4 x, and y - 3 = -3/4 ( x - (-4)), which on solving for y gives you y = -3/4 x. **
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RESPONSE --> I didn't take the positive and the negative of the square root like I should have so I only have the positive values, but my implicit differentian was correct as was my positive Ytan line.
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