course mth173 ?????C???????assignment #011011. Rules for calculating derivatives of some functions.
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21:29:03 `qNote that there are 9 questions in this assignment. `q001. The most basic functions you studied precalculus were: the power functions y = x^n for various values of n, the exponential function y = e^x, the natural logarithm function y = ln(x), and the sine and cosine functions y = sin(x) and y = cos(x). We have fairly simple rules for finding the derivative functions y ' corresponding to each of these functions. Those rules are as follows: If y = x^n for any n except 0, then y ' = n x^(n-1). If y = e^x then y ' = e^x (that's right, the rate of change of this basic exponential function is identical to the value of the function). If y = ln(x) then y ' = 1/x. If y = sin(x) then y ' = cos(x). If y = cos(x) then y ' = - sin(x). There are also some rules for calculating the derivatives of combined functions like the product function x^5 * sin(x), the quotient function e^x / cos(x), or the composite function sin ( x^5). We will see these rules later, but for the present we will mention one easy rule, that if we multiply one of these functions by some constant number the derivative function will be the derivative of that function multiply by the same constant number. Thus for example, since the derivative of sin(x) is cos(x), the derivative of 5 sin(x) is 5 cos(x); or since the derivative of ln(x) is 1 / x, the derivative of -4 ln(x) is -4 (1/x) = -4 / x. Using these rules, find the derivatives of the functions y = -3 e^x, y = .02 ln(x), y = 7 x^3, y = sin(x) / 5.
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RESPONSE --> y= -3 e^x then y' = -3 e^x y= .02 ln (x) then y'= .02 (1/ x) = .02 / x y= 7x^3 then y'= 7 (3x^(3-1)) = 21x^2 y= sin(x)/5 then y' = cos(x) / 5 confidence assessment: 2
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21:29:48 The 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.
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RESPONSE --> okay self critique assessment: 3
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21:34:00 `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?
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RESPONSE --> since y= 5 * ln (t) the derivatve is y'=5 (1/t ) y' = 5/t and evaluated for t=10 y' = 5/(10) y' = .5 So the rate of change at t=10 is .5cm/s confidence assessment: 2
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21:35:25 The 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.
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RESPONSE --> Yes assuming that y is in cm and t is in sec. but this could be any set of units. self critique assessment: 2
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21:39:56 `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?
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RESPONSE --> The depth function is y= e^t/10 since the derivative of an exponetial fucntion is the same e^t of the function then the derivative is the same as the depth function. y'= e^t / 10 y'= e^(2) / 10 y'= (7.389) / 10 y'= .7389 cm /sec. confidence assessment: 2
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21:40:29 The 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.
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RESPONSE --> okay self critique assessment: 3
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21:46:44 `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?
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RESPONSE --> The altitude function is y= 12 t^3 since 12 is a multiplier it is carried over to the derivative y'= 12 3t^(3-1) = 36t^2 y' = 36 (15)^2 y' = 36 (225) y' = 8100 cm / sec. Again the cm/ sec. could be anything of linear distance divided by time units. confidence assessment: 3
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21:47:06 The 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.
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RESPONSE --> okay self critique assessment: 3
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21:57:42 `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 what rate is position changing when t = 0, when t = `pi/2, and when t = 4?
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RESPONSE --> y = .35 sin (t) the multiplier is .35 and is carried over to the derivative y'= .35 cos(t) y'= .35 cos(0) y'= .35 (1) y'= .35 y'= .35 cos('pi/2) y'= .35 (0) y'= 0 y'= .35 cos(4) y'= .35 (-.785) y'= -.229 confidence assessment: 2
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22:00:50 The 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).
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RESPONSE --> The positive and negative slopes indicate the rate of change of the pendulum from the lowest point (bdc), assuming that left is negative then -.23 the pendulum will be moving at .23cm/sec to the left. self critique assessment: 2
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22:14:04 `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?
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RESPONSE --> With the function y= 4x^3 - 7x^2 + 6x we can break the function down into the functions: 4x^3, 7x^2, 6x the derivatives are 12x^2, 14x, and 6x will cancel itself out when derived. so y'= 12x^2- 14x doing the same process for the other equations we get: y'= 4cos(x) + 8/x and y'= 5e^x - 15x^4 confidence assessment: 2
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22:28:44 Since 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.
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RESPONSE --> can 6x be derived, if so how? When plugging in the (x+ 'dx) to make the equation = 'dy/'dx it canceled it self out in the division and then 6-6 = 0. Or is it just known that the derivative of nx = n? Also on the last one I didn't see that the exponent was negative in 3x^-5 therefor my y' for that answer was of -15x^4 where it should have been -15x^-6. other than that the equations seem to match up. self critique assessment: 2
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22:39:01 `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?
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RESPONSE --> for the function y= x^3 * sin (x) y' = (3x^2)(sin(x)) + (cos(x))(x^3) y= e^t cos(t) y'= (e^t sin(1/t) + (cos(t) e^t) y= ln(z) * z^-3 y'= (1/z)(z^-3) + (-3z^-4)(ln(z)) confidence assessment: 2
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22:47:03 The 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)).
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RESPONSE --> The second and third equations could have been simplified. Also the second equation I said that the derivative of cos(t) was sin (1/t) where it was -sin(t), upon reworking I get the same answer as you. self critique assessment: 2
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23:03:28 `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)?
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RESPONSE --> y = e^t / t^5 where f = e^t and g = t^5 y' = [ f ' g - g ' f ] / g^2 y'= ((e^t)(t^5) - (5t^4)(e^t)) / (t^5)^2 y'= ((t^5) - (5t^4)) / (t^10) y' = (t^-5) - (5t^4) y= sin(x) / cos(x) y'= ((cos(x))(cos(x))) - ((-sin(x))(sin(x))) / (cos(x) ^2) y'= (cos(x))^2 - ((-sin(x))(sin(x))) / (cos(x) ^2) y'= -sin(x) sin(x) y= ln(x) / sin (x) y'= ((1/x)(sin(x))) - ((cos(x)(ln(x))) / (sin (x)^2) confidence assessment: 2
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23:14:59 The 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.
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RESPONSE --> (e^t * t^5 - e^t * 5 t^4) / (t^5)^2 The e^t would not cancel out due to the fact that it is multiplied by different amounts first also reworking I get the correct function On the secind problem I assume that since we are subtracting a -sin then this becomes sin and when multiplied by another sin you get the sin(x) ^2 thus making the pythagorean identity applicable. self critique assessment: 2
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23:17:12 The 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.
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RESPONSE --> okay self critique assessment:
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23:51:55 `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.
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RESPONSE --> y=4 ln(x) / sin(x) - sin(x) * cos(x) 4ln(x) / sin(x) minus sin(x) *cos(x) y='((f'g - g'f) /g^2)) - (f'g+g'f) y'= ((4/x)(sin(x) - (cos(x)(4ln(x)) /(4ln(x)^2) - (cos(x)(cos(x)) + (-sin(x))(sin(x)) y'= (4sin(x)/x) - (cos(x)(4ln(x)) /(4ln(x)^2) - (cos(x)^2) + -(sin(x))^2 y= 3e^t / t + 6ln(t) y'= ((f'g - g'f) /g^2)) + (f') y'= (((3e^t)(t))-((t)(3e^t))/ (t^2)) + (6/t) y'= (0 / t^2) + 6/t y'= 6/t y= -5t^5 / ln(t) + sin(t)/5 y'= ((f'g - g'f) /g^2)) + ((f'g - g'f) /g^2)) y' ((-25t^-6)(ln(t))- (1/t)(-5t^5)) / (ln(t)^2)) + ((cos(t))(5) - ((5 (sinI(t)) / (5)^2 y'= ((-25t^-6)(ln(t)) - (-5t^5/t)) / (ln(t) ^2) + (5 (cos(t) -son(t)) /25 y'= ((-25t^-6) -(5t^5)) + (.2 (cos(t) -sin(t))) confidence assessment: 1
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23:57:56 Since 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.
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RESPONSE --> I understand the rules of finding the derivative and when the derivatives are added multiplied etc. however when all the steps where combined and I tried to simplify I got mixed up in all of the parenthesis. In the future I will need to individually break down each of the equations into there different parts and derive those then combine the derivatives on a smaller scale to obtain the total derivative of the equation. self critique assessment: 2
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