#$&* course Mth 173 10/14 1:20
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Given Solution: `aThe derivative of y = -3 e^x is -3 times the derivative of y = e^x. Since by the given rules the derivative of e^x is e^x, the derivative of y = - 3 e^x is y ' = - 3 e^x. The derivative of y = .02 ln(x) is .02 times the derivative of y = ln(x). Since the derivative of ln(x) is 1 / x, the derivative of y = .02 ln(x) is y ' = .02 * 1 / x = .02 / x. The derivative of y = 7 x^3 is 7 times the derivative of x^3. Since the derivative of x^n is n x^(n-1), the derivative of x^3 is 3 x^(3-1), or 3 x^3. The derivative of y = 7 x^3 is therefore y ' = 7 ( 3 x^2) = 21 x^2. The derivative of y = sin(x) / 5 is 1/5 the derivative of sin(x). The derivative of sin(x), according to the rules given above, is cos(x). Thus the derivative of y = sin(x) / 5 is y ' = 1/5 cos(x), or y ' = cos(x) / 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q002. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = 5 * ln(t), then at what rate is water rising in the container when t = 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given the depth function y = 5 * ln(t) depth unit. The rate at which the water is rising in the container as a function of time will be obtained by differentiating the depth function with respect to time. Thus the rate at which water rises = y’ = 5 * ( 1/t) = 5/t depth unit/time unit . Now the rate at which water rises at t = 10 will be y’(10) = 5/10 = 1/2 depth unit/time unit. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of ln(t) is 1 / t, we have rate = y ' = 5 * 1 / t = 5 / t. Since the rate is y ' = 5 / t, when t = 10 the water is rising at rate y ' = 5 / 10 = .5. If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .5 cm/sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q003. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = e^t / 10, then at what rate is water rising in the container when t = 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given the depth function y = e^t / 10 depth unit. The rate at which the water is rising in the container as a function of time will be obtained by differentiating the depth function with respect to time. Thus the rate at which water rises = y’ = e^t / 10 depth unit/time unit . Now the rate at which water rises at t = 2 will be y’(2) = e^2/10 = depth unit/time unit. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe time rate at which water is rising is the derivative y ' = dy / dt of the depth function y. Since the derivative of e^t is e^t, we have rate = y ' = e^t / 10. Since the rate is y ' = e^t / 10, when t = 2 the water is rising at rate y ' = e^2 / 10 = .73, approx. If y is depth in cm and t is clock time in seconds, then the rate is y ' = dy / dt = .73 cm/sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q004. If the altitude of a certain rocket is given as a function of clock time t by y = 12 * t^3, then what function gives the rate of altitude change, and at what rate is the altitude changing when t = 15? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given altitude function y = 12 * t^3 height unit The rate of change of altitude is given by differentiating the rate function with time. Thus the rate of change of altitude function y’ = 12 * ( 3 * t^2) = 36 * t^2 depth unit/ time unit. Thus the value of the rate of change of altitude at t = 15 is given by y’(15) = 36 * (15)^2 = 8100 depth unit / time unit confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe time rate at which altitude is changing is the derivative y ' = dy / dt of the depth function y. Since the derivative of t^3 is 3 t^2, we have rate = y ' = 12 * (3 t^2) = 36 t^2. Since the rate is y ' = 36 t^2, when t = 15 the altitude is changing at rate y ' = 36 * 15^2 = 8100, approx. If y is altitude in feet and t is clock time in seconds, then the rate is y ' = dy / dt = 8100 ft/sec. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q005. If the position of a certain pendulum is given relative to its equilibrium position by the function y = .35 sin(t), then what function gives the corresponding rate of position change, and at what rate is position changing when t = 0, when t = `pi/2, and when t = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The position of a pendulum is given by y = .35 sin(t). The function that gives the corresponding rate of position change with respect to the time is given by differentiating the y function of the pendulum. Thus y’ = dy / dt = .35 * cos(t). Thus the rate of position change when t = pi/2 and t = 4 will be calculated as y’(pi/2) = .35 * cos(pi/2) = .35 * 0 = 0 and y’(4) = .35 * cos(4) = - .35 * .654 approx = -.2289 approx confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe time rate at which position is changing is the derivative y ' = dy / dt of the position function y. Since the derivative of sin(t) is cos(t), we have rate = y ' = .35 cos(t). Since the rate is y ' = .35 cos(t), When t = 0 the position is changing at rate y ' = .35 cos(0) = .35. When t = `pi/2 the position is changing at rate y ' = .35 cos(`pi/2) = 0. When t = 4 the position is changing at rate y ' = .35 cos(4) = -.23. If y is position in cm and t is clock time in seconds, then the rates are .35 cm/s (motion in the positive direction), -.35 cm/s (motion in the negative direction), and -.23 cm/s (motion in the negative direction but not quite as fast). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q006. Another rule is not too surprising: The derivative of the sum of two functions is the sum of the derivatives of these functions. What are the derivatives of the functions y = 4 x^3 - 7 x^2 + 6 x, y = 4 sin(x) + 8 ln(x), and y = 5 e^x - 3 x^-5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given y = 4 x^3 - 7 x^2 + 6 x The derivative of x^3 is 3x^2 and that of x^2 is 2x and x is 1, thus the derivative y’ of the function y is y’ = 12*x^2 - 14x + 6. Given y = 4 sin(x) + 8 ln(x) The derivative of sin(x) is cos(x) and that of ln(x) is 1/x, thus the derivative y’ of the function y is y’ = 4 cos(x) + 8/x. Given y = 5e^x - 3 x^-5 The derivative of e^x is e^x and that of x^-5 is -5x^-6, thus the derivative y’ of the function y is y’ = 5e^x + 15 x^-6. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince y = 4 x^3 - 7 x^2 + 6 x is the sum of the functions 4 x^3, -7 x^2 and 6x, whose derivatives are 12 x^2, -14 x and 6, respectively, we see that y ' is the sum of these derivatives: y ' = 12 x^2 - 14 x + 6. Since y = 4 sin(x) + 8 ln(x) is the sum of the functions 4 sin(x) and 8 ln(x), whose derivatives are , respectively, 4 cos(x) and 8 / x, we see that y ' is the sum of these derivatives: y ' = 4 cos(x) + 8 / x Since y = 5 e^x - 3 x^-5 is the sum of the functions 5 e^x and 3 x^-5, whose derivatives are, respectively, 5 e^x and -15 x^-6, we see that y ' is the sum of these derivatives: y ' = 5 e^x + 15 x^-6. Note that the derivative of x^-4, where n = -4, is n x^(n-1) = -4 x^(-4-1) = -4 x^-5. STUDENT QUESTION I understand how to take the derivative but I am confused by what is meant by the statement of “The derivative of the sum of two functions is the sum of the derivatives of these functions” What does this mean? INSTRUCTOR RESPONSE You used this property without really thinking about it when you found the derivative of 4 sin(x) + 8 ln(x). 4 sin(x) + 8 ln(x) is the sum of two functions, 4 sin(x) and 8 ln(x). Each function has a derivative: • (4 sin(x)) ' = 4 cos(x) and • (8 ln(x)) ' = 8 * 1/x = 8/x. The stated property assures you that the derivative of the sum 4 sin(x) + 8 ln(x) is the sum 4 cos(x) + 8 / x of the two derivatives. You also used the property in each of the other solutions, again really with even noticing that you had used it. It's pretty automatic. However the rules for product and quotient functions are not what you would at first expect them to be, as you will see in the next question. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q007. The rule for the product of two functions is a bit surprising: The derivative of the product f * g of two functions is f ' * g + g ' * f. What are the derivatives of the functions y = x^3 * sin(x), y = e^t cos(t), and y = ln(z) * z^-3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y = x^3 * sin(x) Here x^3 = f(x) and sin(x) = g(x) The derivative of the function x^3 is 3 x^2 and that of sin(x) is cos(x). Thus the derivative of the function y, y’ = f(x) * g’(x) + g(x) * f’(x). Thus y’ = x^3 * cos(x) + sin(x) * (3*x^2) Given function y = e^t * cos(t) The derivative of the function e^t is e^t and that of the function cos(t) is - sin(t) Thus similarly the derivative of the function y’ = e^t * - sin(t) + cos(t) * e^t And finally the function y = ln(z) * z^-3 The derivative of the function ln(z) is 1/z and that of z^-3 is -3 z^-4 Thus the derivative of the function y is y’ = ln(z) * ( -3 z^-4) + ( z^-3) * (1/z) = z^-4 ( -3 ln(z) + 1 ) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe derivative of y = x^3 * sin(x), which is of form f * g if f = x^3 and g = sin(x), is f ' g + g ' f = (x^3) ' sin(x) + x^3 (sin(x)) ' = 3x^2 sin(x) + x^3 cos(x). The derivative of y = e^t cos(t), which is of form f * g if f = e^t and g = cos(t), is f ' g + g ' f = (e^t) ' cos(t) + e^t (cos(t) ) ' = e^t cos(t) + e^t (-sin(t)) = e^t [ cos(t) - sin(t) ]. The derivative of y = ln(z) * z^-3, which is of form f * g if f = ln(z) and g = z^-3, is f ' g + g ' f = (ln(z)) ' z^-3 +ln(z) ( z^-3) ' = 1/z * z^-3 + ln(z) * (-3 z^-4) = z^-4 - 3 ln(z) * z^-4 = z^-4 (1 - 3 ln(z)). STUDENT COMMENT I should have simplified my answers more. INSTRUCTOR RESPONSE Typically students don't really learn their algebra until they have to put their derivatives into standard form so they can compare their answers with the answers in the back of the book. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q008. The rule for the quotient of two functions is perhaps even more surprising: The derivative of the quotient f / g of two functions is [ f ' g - g ' f ] / g^2. What are the derivatives of the functions y = e^t / t^5, y = sin(x) / cos(x) and y = ln(x) / sin(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Given function y(t) = e^t / t^5, here f(t) = e^t and f’(t) = e^t g(t) = t^5 and g’(t) = 5*t^4. Thus the derivative of the function y(t), y’(t) will thus be y’(t) = ( g(t)*f’(t) - f(t)*g’(t) ) / g^2(t) = ( t^5*e^t - e^t*(5t^4)) / t^10 y’(t) = [ e^t * t^4 ( t - 5)]/t^10 = ( e^t ( t - 5)) / t^6 Given function y = sin(x) / cos(x). f(x) = sin(x) and g(x) = cos(x) And thus f’(x) = cos(x) and g’(x) = - sin(x) y’ = dy/dx = [ cos(x)*cos(x) - sin(x)*-sin(x)] / cos^2(x) = ( cos^2(x) + sin^2(x)) / cos^2(x) = 1 / cos^2(x) = sec^2(x) Given function y = ln(x) / sin(x) f(x) = ln(x) and g(x) = sin(x) where f’(x) = 1/x and g’(x) = cos(x) Thus f’(x) = ( sin(x)*(1/x) - ln(x)cos(x)) / sin^2(x) confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aThe derivative of y = e^t / t^5, which is of form f / g if f = e^t and g = t^5, is (f ' g - g ' f) / g^2= ( (e^t) ' t^5 - e^t (t^5) ' ) / (t^5)^2 = (e^t * t^5 - e^t * 5 t^4) / (t^5)^2 = t^4 * e^t ( t - 5) / t^10 = e^t (t-5) / t^6.. The derivative of y = sin(x) / cos(x), which is of form f / g if f = sin(x) and g = cos(x), is (f ' g + g ' f) / g^2 =( (sin(x)) ' cos(x) - sin(x) (cos(x)) ' ) / (cos(x))^2 ' = (cos(x) * cos(x) - sin(x) * -sin(x) ) / (cos(x))^2 = ( (cos(x))^2 + (sin(x))^2 ) / (cos(x))^2 = 1 / cos(x)^2. Note that we have used the Pythagorean identity (sin(x))^2 + (cos(x))^2 = 1. The derivative of y = ln(x) / sin(x), which is of form f / g if f = ln(x) and g = sin(x), is (f ' g + g ' f) / g^2 =( (ln(x)) ' sin(x) - ln(x) (sin(x)) ' ) / (sin(x))^2 = (sin(x) * 1/x - ln(x) * cos(x) ) / (sin(x))^2 = ( sin(x) / x - ln(x) cos(x) ) / (sin(x))^2 = 1 / ( x sin(x)) - ln(x) cos(x) / (sin(x))^2. Further simplification using the tangent function is possible, but the answer here will be left in terms of the sine and cosine functions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q009. Combining the above rules find the derivatives of the following functions: y =4 ln(x) / sin(x) - sin(x) * cos(x); y = 3 e^t / t + 6 ln(t), y = -5 t^5 / ln(t) + sin(t) / 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 4 ln(x) / sin(x) - sin(x) * cos(x) Differentiation of ln(x) is 1/x, sin(x) is cos(x) and that of cos(x) is - sin(x). On using all the rules in the previous question we get the differential of the function y(x), y’(x) will be y’(x) = 4 ( sin(x) * ( 1/x ) - ln(x) * cos(x) ) / sin^2(x) - ( sin(x)*-sin(x) + cos(x)*(cos(x)) y’(x) = 4 ( sin(x) / x - ln(x) * cos(x) ) / sin^2(x) + sin^2(x) - cos^2(x) Next function y(t) = -5 t^5 / ln(t) + sin(t) / 5 The function ln(t) has a derivative of 1/t and sin(t) will have a derivative cos(t) The differential of the given function y’(t) will thus be y’(t) = -5 * ( ln(t) * 5t^4 - ( t^5 * 1/t ) ) / ln^2(t) + (1/5) * cos(t) y’(t) = -5 * ( ln(t) * 5t^4 - t^4 ) / ln^2(t) + cos(t)/5 y’(t) = -5*t^4( 5*ln(t) - 1 ) / ln^2(t) + cos(t)/5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince the derivative of sin(x) / ln(x) is 1 / ( x sin(x)) + ln(x) cos(x) / (sin(x))^2 ), as just seen, and the derivative of sin(x) * cos(x) is easily seen by the product rule to be -(sin(x))^2 + (cos(x))^2, we see that the derivative of y = 4 sin(x) / ln(x) - sin(x) * cos(x) is y ' = 4 [ 1 / ( x sin(x) - ln(x) cos(x) / (sin(x))^2 ] - ( -(sin(x))^2 + (cos(x))^2 ) = 4 / ( x sin(x) ) - 4 ln(x) cos(x) / ( sin(x))^2 + (sin(x))^2 - (cos(x))^2. Further rearrangement is possible but will not be done here. The derivative of 3 e^t / t is found by the quotient rule to be ( 3 e^t * t - 3 e^t * 1 ) / t^2 = 3 e^t ( t - 1) / t^2, the derivative of 6 ln(t) is 6 / t, so the derivative of y = 3 e^t / t + 6 ln(t) is therefore y ' = 3 e^t ( t - 1) / t^2 + 6 / t. Since the derivative of -5 t^5 / ln(t) is found by the quotient rule to be ( -25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2, and the derivative of sin(t) / 5 is cos(t) / 5, we see that the derivative of y = -5 t^5 / ln(t) + sin(t) / 5 is y ' = (-25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2 + cos(t) / 5 = -25 t^4 ln(t) + 5 t^4 / (ln(t))^2 + cos(t) / 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q009. Combining the above rules find the derivatives of the following functions: y =4 ln(x) / sin(x) - sin(x) * cos(x); y = 3 e^t / t + 6 ln(t), y = -5 t^5 / ln(t) + sin(t) / 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 4 ln(x) / sin(x) - sin(x) * cos(x) Differentiation of ln(x) is 1/x, sin(x) is cos(x) and that of cos(x) is - sin(x). On using all the rules in the previous question we get the differential of the function y(x), y’(x) will be y’(x) = 4 ( sin(x) * ( 1/x ) - ln(x) * cos(x) ) / sin^2(x) - ( sin(x)*-sin(x) + cos(x)*(cos(x)) y’(x) = 4 ( sin(x) / x - ln(x) * cos(x) ) / sin^2(x) + sin^2(x) - cos^2(x) Next function y(t) = -5 t^5 / ln(t) + sin(t) / 5 The function ln(t) has a derivative of 1/t and sin(t) will have a derivative cos(t) The differential of the given function y’(t) will thus be y’(t) = -5 * ( ln(t) * 5t^4 - ( t^5 * 1/t ) ) / ln^2(t) + (1/5) * cos(t) y’(t) = -5 * ( ln(t) * 5t^4 - t^4 ) / ln^2(t) + cos(t)/5 y’(t) = -5*t^4( 5*ln(t) - 1 ) / ln^2(t) + cos(t)/5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince the derivative of sin(x) / ln(x) is 1 / ( x sin(x)) + ln(x) cos(x) / (sin(x))^2 ), as just seen, and the derivative of sin(x) * cos(x) is easily seen by the product rule to be -(sin(x))^2 + (cos(x))^2, we see that the derivative of y = 4 sin(x) / ln(x) - sin(x) * cos(x) is y ' = 4 [ 1 / ( x sin(x) - ln(x) cos(x) / (sin(x))^2 ] - ( -(sin(x))^2 + (cos(x))^2 ) = 4 / ( x sin(x) ) - 4 ln(x) cos(x) / ( sin(x))^2 + (sin(x))^2 - (cos(x))^2. Further rearrangement is possible but will not be done here. The derivative of 3 e^t / t is found by the quotient rule to be ( 3 e^t * t - 3 e^t * 1 ) / t^2 = 3 e^t ( t - 1) / t^2, the derivative of 6 ln(t) is 6 / t, so the derivative of y = 3 e^t / t + 6 ln(t) is therefore y ' = 3 e^t ( t - 1) / t^2 + 6 / t. Since the derivative of -5 t^5 / ln(t) is found by the quotient rule to be ( -25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2, and the derivative of sin(t) / 5 is cos(t) / 5, we see that the derivative of y = -5 t^5 / ln(t) + sin(t) / 5 is y ' = (-25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2 + cos(t) / 5 = -25 t^4 ln(t) + 5 t^4 / (ln(t))^2 + cos(t) / 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&! ********************************************* Question: `q009. Combining the above rules find the derivatives of the following functions: y =4 ln(x) / sin(x) - sin(x) * cos(x); y = 3 e^t / t + 6 ln(t), y = -5 t^5 / ln(t) + sin(t) / 5. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 4 ln(x) / sin(x) - sin(x) * cos(x) Differentiation of ln(x) is 1/x, sin(x) is cos(x) and that of cos(x) is - sin(x). On using all the rules in the previous question we get the differential of the function y(x), y’(x) will be y’(x) = 4 ( sin(x) * ( 1/x ) - ln(x) * cos(x) ) / sin^2(x) - ( sin(x)*-sin(x) + cos(x)*(cos(x)) y’(x) = 4 ( sin(x) / x - ln(x) * cos(x) ) / sin^2(x) + sin^2(x) - cos^2(x) Next function y(t) = -5 t^5 / ln(t) + sin(t) / 5 The function ln(t) has a derivative of 1/t and sin(t) will have a derivative cos(t) The differential of the given function y’(t) will thus be y’(t) = -5 * ( ln(t) * 5t^4 - ( t^5 * 1/t ) ) / ln^2(t) + (1/5) * cos(t) y’(t) = -5 * ( ln(t) * 5t^4 - t^4 ) / ln^2(t) + cos(t)/5 y’(t) = -5*t^4( 5*ln(t) - 1 ) / ln^2(t) + cos(t)/5 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `aSince the derivative of sin(x) / ln(x) is 1 / ( x sin(x)) + ln(x) cos(x) / (sin(x))^2 ), as just seen, and the derivative of sin(x) * cos(x) is easily seen by the product rule to be -(sin(x))^2 + (cos(x))^2, we see that the derivative of y = 4 sin(x) / ln(x) - sin(x) * cos(x) is y ' = 4 [ 1 / ( x sin(x) - ln(x) cos(x) / (sin(x))^2 ] - ( -(sin(x))^2 + (cos(x))^2 ) = 4 / ( x sin(x) ) - 4 ln(x) cos(x) / ( sin(x))^2 + (sin(x))^2 - (cos(x))^2. Further rearrangement is possible but will not be done here. The derivative of 3 e^t / t is found by the quotient rule to be ( 3 e^t * t - 3 e^t * 1 ) / t^2 = 3 e^t ( t - 1) / t^2, the derivative of 6 ln(t) is 6 / t, so the derivative of y = 3 e^t / t + 6 ln(t) is therefore y ' = 3 e^t ( t - 1) / t^2 + 6 / t. Since the derivative of -5 t^5 / ln(t) is found by the quotient rule to be ( -25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2, and the derivative of sin(t) / 5 is cos(t) / 5, we see that the derivative of y = -5 t^5 / ln(t) + sin(t) / 5 is y ' = (-25 t^4 ln(t) - (-5 t^5 ) * ( 1 / t ) ) / (ln(t))^2 + cos(t) / 5 = -25 t^4 ln(t) + 5 t^4 / (ln(t))^2 + cos(t) / 5. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK #*&!#*&!