course Mth 173 4/3 12 Question: `qQuery problem 3.6.12 (3d edition 3.6.44). was 4.6.12 derivative of e^( ln(x) + 1)
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qWhy is in easier to calculate the derivative of this function if you simplify the function first? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If you simplify this function in particular, and many others, the simplified version usually rules has some factors that cancel out and plus you can make the function less of a composite function so it is easier to look at. confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The simplified factors partially cancel out. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery problem 3.6.43 (3d edition 3.6.44) was 4.6.30 y = ln(x) at x = 1 What is the equation of the tangent line at the given point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 1(x) – 1 The tangent line has a slope of 1 like the original, and passes through y = -1 since the tangent line passes through (0 , 1). confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qWhat are your approximations to ln(1.1) and ln(2), based on the tangent line? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 1.1 would boil down to equaling (x), so just plug 1.1 into function as x and get .1. Do the same for 2 and get 1. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** In terms of the graph, explain whether your approximations are larger or smaller than the true values. 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 ** Since the graph of y = ln(x) is always concave downward, it will always fall below its tangent-line approximation. ** ** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The above problem gives the functions ln(1.1) and ln(2), not ln(1.2). ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `qQuery problem 3.7.16 (was 3.7.9 was 4.7.6) e^(x^2) + ln y = 0 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(x) = e^(x^2) + ln y = 0 F’(x) = 2x(e^2) + 1/y(y’) I can’t tell if I’m finished but I don’t think I am. confidence rating #$&* 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aWhen 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): y ' = 2x e^(x^2) * y ------------------------------------------------ Self-critique rating #$&* 1 ********************************************* Question: `qQuery problem 3.7.34 (3d edition 3.7.26) was 4.7.18 circle x^2+y^2=25 What are the equations of the tangent lines to the circle at the points where x = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: x^2 + y^2 = 25 (4)^2 + y^2 = 25 y = 3 , -3 Points on the circle are at (4, 3) and (4, -3). Since the origin is at (0, 0) then there are 2 slopes, one being positive and the other negative. y = 4/3x – 8.3 is the equation of a tangent line. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): The problem asks for two tangent lines with the other equation having a slope of – 4/3. ------------------------------------------------ Self-critique rating #$&* 3