#$&* course MTH 173 11/14 15:10
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Given Solution: `a** The function is `sqrt( (x^2 * 5^x)^3 ) = (x^2 * 5^x)^(3/2). This is of form f(g(x)) with g(x) = x^2 * 5^x and f(z) = z^(3/2). Thus when you substitute you get f(g(x)) = g(x)^(3/2) = (x^2 * 5^x)^(3/2). (x^2 * 5^x) ' = (x^2)' * 5^x + x^2 * (5^x) ' = 2x * 5^x + x^2 ln 5 * 5^x = (2x + x^2 ln 5) * 5^x. `sqrt(z^3) = z^(3/2), so using w(x) = f(g(x)) with f(z) = z^(3/2) and g(x) = x^2 * 5^x we get w ' = (2x + x^2 ln 5) * 5^x * [3/2 (x^2 * 5^x)^(1/2)] = 3/2 (2x + x^2 ln 5) * | x | * 5^(1/2 x). Note that sqrt(x^2) is | x |, not just x, since the square root must be positive and x might not be. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.4.26 was 3.4.28 (3d edition 3.4.19) (was 4.4.20) derivative of 2^(5t-1). What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y(t) = 2^(5t - 1) This is a compound function with inner function is 5t - 1 and the outer function is 2^x. Using chain rule of differentiation, we get y’(t) = (2^(5t -1))*(ln(2))*(5t - 1)’ = (2^(5t - 1)*(5ln(2)) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThis function is a composite. The inner function is g(x)=5t-1 and the outer function is f(z)=2^z. f'(z)=ln(2) * 2^z. g ' (x)=5 so (f(g(t)) ' = g ' (t)f ' (g(t))= 5 ln(2) * 2^(5t-1). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q**** Query 3.4.67 was 3.4.68 (3d edition 3.4.56) y = k (x), y ' (1) = 2. What is the derivative of k(2x) when x = 1/2? What is the derivative of k(x+1) when x = 0? {]What is the derivative of k(x/4) when x = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Question submitted via text_19 assignment confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We apply the Chain Rule: ( k(2x) ) ' = (2x) ' * k'(2x) = 2 k ' (2x). When x = 1/2 we have 2x = 1. k ' (1) = y ' (1) = 2 so when x = 1/2 ( k(2x) ) ' = 2 k ' (2 * 1/2) = 2 * k'(1) = 2 * 2 = 4. (k(x+1)) ' = (x+1)' k ' (x+1) = k ' (x+1) so when x = 0 we have (k(x+1) ) ' = k ' (x+1) = k ' (1) = 2 (k(x/4)) ' = (x/4)' k'(x/4) = 1/4 * k'(x/4) so when x = 4 we have (k(x/4))' = 1/4 * k'(x/4) = 1/4 k'(4/4) = 1/4 * 2 = 1/2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 3.4.81 (3d edition 3.4.68). Q = Q0 e^(-t/(RC)). I = dQ/dt. Show that Q(t) and I(t) both have the same time constant. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given that Q = Q0 e^(-t/(RC)) and I = dQ/dt The Q function is an exponential functional. The function is a compound function, the inner function is -t/Rc and the outer function is Q0*(e^t). Using the chain rule of differentiation, we obtain Q’ = Q0(e^(-t/RC))’ = Q0(e^(-t/RC))*(-t/RC)’ = Q0(e^(-t/RC))(1/RC) Thus we obtain I = dO/dt = Q’(t) = (Q0/RC)(e^(-t/RC)) And Q(t) = Q0 e^(-t/RC) We thus see that both Q(t) and I(t) share the same time constant. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We use the Chain Rule. (e^(-t/(RC)))' = (-t/(RC))' * e^(-t/(RC)) = -1/(RC) * e^(-t/(RC)). So dQ/dt = -Q0/(RC) * e^(-t/(RC)). Both functions are equal to a constant factor multiplied by e^(-t/(RC)). The time constant for both functions is therefore identical, and equal to RC. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.5.5 (unchanged since 3d edition) (formerly 4.5.6). What is the derivative of sin(3x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y = sin(3x) The given function is a compound function, the inner function is 3x and the outer function is sin(t) Assume that f(x) = 3x and g(t) = sin(t) Thus y = g(f(x)). Using the chain rule of differentiation, we find the derivative of the function y Thus y’ = (g(f(x))’ = g’(f(x))*f’(x) g’(t) = cos(t) and f’(x) = 3 thus g’(f(x)) = cos(3x) thus y’ = cos(3x)*3 the derivative of sin(3x), thus would be equal to 3cos(3x). confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** sin(3x) is the composite of g(x) = 3x, which is the 'inner' function (the first function that operates on the variable x) and the 'outer' function f(z) = sin(z). Thus f(g(x)) = sin(g(x)) = sin(3x). The derivative is (f (g(x) ) ' = g ' (x) * f' ( g(x) ). g ' (x) = (3x) ' = 3 * x ' = 3 ', and f ' (z) = (sin(z) ) ' = cos(z). So the derivative is [ sin(3x) ] ' = ( f(g(x) ) ' = g ' (x) * f ' (g(x) ) = 3 * cos(3x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.5.50 was 3.5.48 (3d edition 3.5.50) (formerly 4.5.36). Give the equations of the tangent lines to graph of y = sin(x) at x = 0 and at `pi/3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y = sin(x) The derivative of the function will thus be y’ = cos(x) Equation of the tangent line at x = 0 will have a slope equal to the derivative of the function at x = 0. Thus slope of the line = y’(0) = cos(0) = 1 Thus value of the function at x = 0 will be y(0) = sin(0) = 0 Thus the function passes through (0,0) Thus the tangent line passes through (0,0) and has a slope of 1. Thus the equation of the tangent line at x = 0 will be y = x. The slope of the tangent line at x = pi/3 will be the value of the derivative at x = pi/3 Thus y’(pi/3) = cos(pi/3) = 1/2 The value of the function at x = pi/3 will thus be equal to y(pi/3) = sin(pi/3) = sqrt(3/4) Thus the tangent line passes through (pi/3 , sqrt(3/4)) and has a slope of ½. Using the slope-point form of equation of line we get (y - sqrt(4/3)) / (x - pi/3) = 1/2 2y - 2sqrt(4/3) = x - pi/3 Thus the equation of the tangent line will be y = ½(x + 2sqrt(4/3) - pi/3) = x/2 + sqrt(4/3) - pi/6 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** At x = 0 we have y = 0 and y ' = cos(0) = 1. The tangent line is therefore the line with slope 1 through (0,0), so the line is y - 0 = 1 ( x - 0) or just y = x. At x = `pi/3 we have y = sin(`pi/3) = `sqrt(3) / 2 and y ' = cos(`pi/3) = .5. Thus the tangent line has slope .5 and passes thru (`pi/3,`sqrt(3)/2), so its equation is y - `sqrt(3)/2 = .5 (x - `pi/3) y = .5 x - `pi/6 + `sqrt(3)/2. Approximating: y - .87 = .5 x - .52. So y = .5 x + .25, approx. Our approximation to sin(`pi/6), based on the first tangent line: The first tangent line is y = x. So the approximation at x = `pi / 6 is y = `pi / 6 = 3.14 / 6 = .52, approximately. Our approximation to sin(`pi/6), based on the second tangent line, is: y = .5 * .52 + .34 = .60. `pi/6 is equidistant from x=0 and x=`pi/3, so we might expect the accuracy to be the same whichever point we use. The actual value of sin(`pi/6) is .5. The approximation based on the tangent line at x = 0 is .52, which is much closer to .5 than the .60 based on the tangent line at x = `pi/3. The reason for this isn't too difficult to see. The slope is changing more quickly around x = `pi/3 than around x = 0. Thus the tangent line will move more rapidly away from the actual function near x = `pi/3 than near x = 0. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery 3.5.34 (3d edition 3.5.40). Der of sin(sin x + cos x) What is the derivative of the given function and how did you find it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y = sin(sin(x) + cos(x)) It is a compounded function, the inner function is sin(x) + cos(x) and the outer function is sin(t) Let us assume g(x) = sin(x) + cos(x), and f(t) = sin(t) Thus we have y = f(g(x)) Using chain rule of differentiation, we have y’ = f’(g(x))*(g’(x)) f’(t) = cos(t) and g’(x) = cos(x) - sin(x) Thus f’(g(x)) = cos(sin(x) + cos(x)) and g’(x) = cos(x) - sin(x) Thus y’ = cos(sin(x) + cos(x))*(cos(x) - sin(x)) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe function y = sin( sin(x) + cos(x) ) is the composite of g(x) = sin(x) + cos(x) and f(z) = sin(z). The derivative of the composite is g ' (x) * f ' (g(x) ). g ' (x) = (sin(x) + cos(x) ) ' = cos(x) - sin(x). f ' (z) = sin(z) ' = cos(z). So g ' (x) * f ' (g(x)) = ( cos(x) - sin(x) ) * cos( sin(x) + cos(x) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery Add comments on any surprises or insights you experienced as a result of this assignment. "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: