course Mth 174 If your solution to stated problem does not match the given solution, you should self-critique per instructions at
.............................................
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 #$&* ok ********************************************* Question: `q002. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = 5 * ln(t), then at what rate is water rising in the container when t = 10? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, you have to take the derivative of the function, which is 5/t, then you would just plug in the t=10 to get ½. confidence rating #$&* 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 #$&* ok ********************************************* Question: `q003. If a container is shaped so that when a certain constant water stream flows into the container, the depth function is y = e^t / 10, then at what rate is water rising in the container when t = 2? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First you have to find the derivative of the function which is e^t / 10, then plug in t=2 to get e^2/10 confidence rating #$&* 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 #$&* ok ********************************************* Question: `q004. If the altitude of a certain rocket is given as a function of clock time t by y = 12 * t^3, then what function gives the rate of altitude change, and at what rate is the altitude changing when t = 15? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: First, I found the derivative of the function, which is 36t^2. By plugging in t=15, then you will get (36)225, which is 8100. confidence rating #$&* 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 #$&* ok ********************************************* Question: `q005. If the position of a certain pendulum is given relative to its equilibrium position by the function y = .35 sin(t), then what function gives the corresponding rate of position change, and what rate is position changing when t = 0, when t = `pi/2, and when t = 4? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of the function is .35cos(t). At t=0, y’=.35cos(0), which equals .35. At t=pi/2, y’=.35cos(pi/2), which equals 0. At t=4, y’=.35cos(4), which equals -.23. confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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 #$&* ok ********************************************* Question: `q006. Another rule is not too surprising: The derivative of the sum of two functions is the sum of the derivatives of these functions. What are the derivatives of the functions y = 4 x^3 - 7 x^2 + 6 x, y = 4 sin(x) + 8 ln(x), and y = 5 e^x - 3 x^-5? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of y = 4 x^3 - 7 x^2 + 6 x is y’=12x^2-14 is y’=12x^2-14x+6. The derivative of y = 4 sin(x) + 8 ln(x) is y’=4cos(x)+8/x The derivative of y = 5 e^x - 3 x^-5 is y’=5e^x+15x^-6 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q007. The rule for the product of two functions is a bit surprising: The derivative of the product f * g of two functions is f ' * g + g ' * f. What are the derivatives of the functions y = x^3 * sin(x), y = e^t cos(t), and y = ln(z) * z^-3? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of y = x^3 * sin(x) is y’=3x^2*sin(x)+cos(x)*x^3= 3x^2sin(x)+cos(x)x^3. The derivative of y = e^t cos(t) is y’=e^t*cos(t)+(-sin(x))*e^t= e^tcos(t)-sin(x)e^t. The derivative of y = ln(z) * z^-3 is y’=(1/z)*z^-3+(-3z^-4)*ln(z)= =(1/z)z^-3-(3z^-4*ln(z)) confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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): ok ------------------------------------------------ Self-critique rating #$&* ok ********************************************* Question: `q008. The rule for the quotient of two functions is perhaps even more surprising: The derivative of the quotient f / g of two functions is [ f ' g - g ' f ] / g^2. What are the derivatives of the functions y = e^t / t^5, y = sin(x) / cos(x) and y = ln(x) / sin(x)? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative of y = e^t / t^5 is y’=(e^t*t^5-5t^4*e^t)/(t^5)^2= e^tt^5-(5t^4e^t)/(t^5)^2= t^4 * e^t ( t - 5) / t^10= e^t *(t-5) / t^6 The derivative of y = sin(x) / cos(x) is y’=cos(x)*cos(x)-sin(x)*-sin(x)/cos(x)^2= cos(x)^2 +sin(x)^2/ cos(x)^2= 1/cos(x)^2 The derivative of y = ln(x) / sin(x) is y’= 1/x*sin(x)-ln(x)*cos(x)/sin(x)^2= sin(x)/x- ln(x)cos(x)/sin(x)^2 confidence rating #$&* 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: confidence rating #$&*2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: `aThe derivative of y = e^t / t^5, which is of form f / g if f = e^t and g = t^5, is (f ' g - g ' f) / g^2= ( (e^t) ' t^5 - e^t (t^5) ' ) / (t^5)^2 = (e^t * t^5 - e^t * 5 t^4) / (t^5)^2 = t^4 * e^t ( t - 5) / t^10 = e^t (t-5) / t^6.. The derivative of y = sin(x) / cos(x), which is of form f / g if f = sin(x) and g = cos(x), is (f ' g + g ' f) / g^2 =( (sin(x)) ' cos(x) - sin(x) (cos(x)) ' ) / (cos(x))^2 ' = (cos(x) * cos(x) - sin(x) * -sin(x) ) / (cos(x))^2 = ( (cos(x))^2 + (sin(x))^2 ) / (cos(x))^2 = 1 / cos(x)^2. Note that we have used the Pythagorean identity (sin(x))^2 + (cos(x))^2 = 1. The derivative of y = ln(x) / sin(x), which is of form f / g if f = ln(x) and g = sin(x), is (f ' g + g ' f) / g^2 =( (ln(x)) ' sin(x) - ln(x) (sin(x)) ' ) / (sin(x))^2 = (sin(x) * 1/x - ln(x) * cos(x) ) / (sin(x))^2 = ( sin(x) / x - ln(x) cos(x) ) / (sin(x))^2 = 1 / ( x sin(x)) - ln(x) cos(x) / (sin(x))^2. Further simplification using the tangent function is possible, but the answer here will be left in terms of the sine and cosine functions. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I think I did the third one a little wrong, but other than that, these were pretty easy for me. ------------------------------------------------ Self-critique rating #$&* 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: The derivative of y =4 ln(x) / sin(x) - sin(x) * cos(x) is y’=4(1 / ( x sin(x)) + ln(x) cos(x)) / (sin(x))^2 )- cos(x)*cos(x)+(sin(x)-sin(x)) which equals 4 / ( x sin(x) ) - 4 ln(x) cos(x) / ( sin(x))^2 + (sin(x))^2 - (cos(x))^2 The derivative of y = 3 e^t / t + 6 ln(t) is (3e^t*t-1*3e^t)/(t^2)+6/x, which equals 3 te^t -3e^t) / t^2 + 6 / t The derivative of = -5 t^5 / ln(t) + sin(t) / 5 is y’=(-25t^4*ln(t)-1/t*-5t^5)/(ln(t)^2 + cos(t)*5-sin(t)*0/25, which equals -25t^4 ln(t)+5t^4 /ln(t)^2+cos(t)/5 confidence rating #$&* 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: 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. "