#$&* course MTH 173 8/7/113pm
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Given Solution: `a** The function is `sqrt( (x^2 * 5^x)^3 ) = (x^2 * 5^x)^(3/2). This is of form f(g(x)) with g(x) = x^2 * 5^x and f(z) = z^(3/2). Thus when you substitute you get f(g(x)) = g(x)^(3/2) = (x^2 * 5^x)^(3/2). (x^2 * 5^x) ' = (x^2)' * 5^x + x^2 * (5^x) ' = 2x * 5^x + x^2 ln 5 * 5^x = (2x + x^2 ln 5) * 5^x. `sqrt(z^3) = z^(3/2), so using w(x) = f(g(x)) with f(z) = z^(3/2) and g(x) = x^2 * 5^x we get w ' = (2x + x^2 ln 5) * 5^x * [3/2 (x^2 * 5^x)^(1/2)] = 3/2 (2x + x^2 ln 5) * | x | * 5^(1/2 x). Note that sqrt(x^2) is | x |, not just x, since the square root must be positive and x might not be. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.4.26 was 3.4.28 (3d edition 3.4.19) (was 4.4.20) derivative of 2^ (5t-1). What is the derivative of the given function? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ y' = ln(2)(5)(2^(5t-1)) &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*: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: &&&&&&&&&&&&&&&&&&&&&&&& ************************ (k(2x))' = 2k'(2x) (k(2x))'(.5) = 2k'(1) = 2(2) = 4 (k(x+1))' = (1)k'(x+1) (k(x+1))'(0) = k'(0+1) = 2 (k(x/4))'= 1/4(k'(x/4)) (k(x/4))'(4) = 1/4(k'(4/4)) = (1/4)(2) = .5 &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** We apply the Chain Rule: ( k(2x) ) ' = (2x) ' * k'(2x) = 2 k(2x). When x = 1/2 we have 2x = 1. k ' (1) = y ' (1) = 2 so when x = 1/2 ( k(2x) ) ' = 2 k(2 * 1/2) = 2 * k'(1) = 2 * 2 = 4. (k(x+1)) ' = (x+1)' k ' (x+1) = k ' (x+1) so when x = 0 we have (k(x+1) ) ' = k ' (x+1) = k ' (1) = 2 (k(x/4)) ' = (x/4)' k'(x/4) = 1/4 * k'(x/4) so when x = 4 we have (k(x/4))' = 1/4 * k'(x/4) = 1/4 k'(4/4) = 1/4 * 2 = 1/2. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery 3.4.81 (3d edition 3.4.68). Q = Q0 e^(-t/(RC)). I = dQ/dt. Show that Q(t) and I(t) both have the same time constant. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ dQ/dt = (Q0)(-1/RC)e^(-t/(RC)) both have the same time constant. &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:OK ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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: &&&&&&&&&&&&&&&&&&&&&&&& ************************ y = sin(3x) y' = (3)cos(3x) &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*:3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** sin(3x) is the composite of g(x) = 3x, which is the 'inner' function (the first function that operates on the variable x) and the 'outer' function f(z) = sin(z). Thus f(g(x)) = sin(g(x)) = sin(3x). The derivative is (f (g(x) ) ' = g ' (x) * f' ( g(x) ). g ' (x) = (3x) ' = 3 * x ' = 3 ', and f ' (z) = (sin(z) ) ' = cos(z). So the derivative is [ sin(3x) ] ' = ( f(g(x) ) ' = g ' (x) * f ' (g(x) ) = 3 * cos(3x). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK ********************************************* Question: `qQuery problem 3.5.50 was 3.5.48 (3d edition 3.5.50) (formerly 4.5.36). Give the equations of the tangent lines to graph of y = sin(x) at x = 0 and at `pi/3 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: &&&&&&&&&&&&&&&&&&&&&&&& ************************ y = sin(x) y' = cos(x) y'(0) = cos(0) = 1 y(0) = 0 so 0 = 0(1) + b b = 0 y = x y'(pi/3) = .5 y(pi/3) = sqrt(3)/2 sqrt(3)/2 = .5(pi/3) + b b =~ .3424 y = .5x + .3424 &&&&&&&&&&&&&&&&&&&&&&&& ************************ confidence rating #$&*: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: &&&&&&&&&&&&&&&&&&&&&&&& ************************ y = sin(sin(x) + cos(x)) y' = (cos(x) - sin(x))(cos(sin(x) + cos(x))) take the derivative of whatever is within sine and multiply it by cosine (the derivative) of whatever was within sine: ex. sin(2x) = 2cos(2x) or (sin(z))' = z'cos(z) &&&&&&&&&&&&&&&&&&&&&&&& ************************ 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" Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!