course Mth 173
.............................................
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 #$&* ********************************************* Question: `q002. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = 5 * ln(t), then at what rate is water rising in the container when t = 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: We are given the depth function and it is known that the rate function is the derivitive of the rate function. We can use the rules stated above to find that y=5*ln(t) = y`=5*1/t so when t=10 we solve 5*1/10=.5 and the rate is .5 with units that are not known. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
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 #$&* 3 ********************************************* Question: `q003. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = e^t / 10, then at what rate is water rising in the container when t = 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of e raised to the T is E raised to the T so in this case y=y` so the derivative of e^t/10 is e^t/10 so replacing T with 2 we solve for a rate of .739 units. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
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 #$&* 3 ********************************************* Question: `q004. If the altitude of a certain rocket is given as a function of clock time t by y = 12 * t^3, then what function gives the rate of altitude change, and at what rate is the altitude changing when t = 15? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Agin the rate is the derivative of the depth function and the derivative of y=12*t^3 is y`=36t^2 when t = 15. Y`=36*15^2=8100 units. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
.............................................
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 #$&* 3 ********************************************* Question: `q005. If the position of a certain pendulum is given relative to its equilibrium position by the function y = .35 sin(t), then what function gives the corresponding rate of position change, and what rate is position changing when t = 0, when t = `pi/2, and when t = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Y=35sin(t) becomes y`=.35cos(t) T=4 y`=.35cos(4)=-.229 T=pi/2 y`=.35cos(pi/2) T=0 y`=.35cos(0)=35 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
.............................................
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): I think I found the rates correctly but I should have indicated what the values actually meant. Example being the negative symbol indicated a downward motion. ------------------------------------------------ Self-critique rating #$&* 3 ********************************************* Question: `q006. Another rule is not too surprising: The derivative of the sum of two functions is the sum of the derivatives of these functions. What are the derivatives of the functions y = 4 x^3 - 7 x^2 + 6 x, y = 4 sin(x) + 8 ln(x), and y = 5 e^x - 3 x^-5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: y = 4 x^3 - 7 x^2 + 6 x y`=12x^2-14x+6 y = 4 sin(x) + 8 ln(x) y`=4 cos(x) + 8/x y = 5 e^x - 3 x^-5 y`=5 e^x -15x^-6 confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 2
.............................................
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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I understand how to take the derivative but I am connfussed by what is ment 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????
.............................................
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)). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I should have simplified my answers more.
.............................................
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. confidence rating #$&* ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): After seeing the solution I think that I should use the model [ f ' g - g ' f ] / g^2 replacing with the corresponding values but does the numerator or the Denominator become f???????????
.............................................
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. "