course MTH 272 02/23, 1200 µsˆÛ•ê³Å Uª·üõ߆öàMÞ ÕþÒ×û²Kˆ¬ù›assignment #003 003. `query 3 Applied Calculus II 02-14-2010
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17:22:54 4.5.10 (was 4.4.10) find the derivative of ln(1-x)^(1/3)
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RESPONSE --> ln(1-x)^(1/3) (1/3) ln (1- x) u=(1-x) du/dx= -1 (1/3)*(1/(1-x))*-1 -1/(3(1-x)) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:23:14 The function is of the form ln(u), so the derivative is 1/u * u', or ln(u) * du/dx. The function u is (1-x)^(3/2). The derivative of this function is u' = du/dx = -1 * 3/2 * (1-x)^(1/2) = -3/2 (1-x)^(1/2). Thus the derivative of the original function is 1/u du/dx = 1 / [(1-x)^(3/2) ] * [-3/2 (1-x)^(1/2)] = -3/2 (1-x)^(1/2) (1-x)^(-3/2) = -3/2 (1-x)^-1 = -3 / [ 2 (1-x) ] ALTERNATIVE SOLUTION: Note that ln(1-x)^(1/3) = 1/3 ln(1-x) The derivative of ln(1-x) is u ' * 1/u with u = 1-x. It follows that u ' = -1 so the derivative of ln(1-x) is -1 * 1/(1-x) = -1/(1-x). The derivative of 1/3 ln(1-x) is therefore 1/3 * -1/(1-x) = -1 / [ 3(1-x) ].**
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 3
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17:26:05 4.5.25 (was 4.4.24) find the derivative of ln( (e^x + e^-x) / 2)
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RESPONSE --> ln( (e^x + e^-x) / 2) (1/(e^x + e^-x)) * (e^x + (-1)e^-x) (e^x - e^-x)/(e^x + e^-x) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:26:31 the derivative of ln(u) is 1/u du/dx; u = (e^x + e^-x)/2 so du/dx = (e^x - e^-x) / 2. The term - e^(-x) came from applying the chain rule to e^-x. The derivative of ln( (e^x + e^-x) / 2) is therefore [(e^x - e^-2) / 2 ] / ] [ (e^x + e^-x) / 2 ] = (e^x - e^-x) / (e^x + e^-x). This expression does not simplify, though it can be expressed in various forms (e.g., (1 - e^-(2x) ) / ( 1 + e^-(2x) ), obtained by dividing both numerator and denominator by e^x.). ALTERNATIVE SOLUTION: ln( (e^x + e^-x) / 2) = ln( (e^x + e^-x) ) - ln(2). the derivative of e^(-x) is - e^(-x) and ln(2) is constant so its derivative is zero. So you get y ' = (e^x - e^-x)/(e^x + e^-x). **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 3
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17:27:48 4.5.30 (was 4.4.30) write log{base 3}(x) in exp form
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RESPONSE --> log {base 3} (x)=y 3^y=x confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:28:21 the exponential form of y = log{base 3}(x) is x = 3^y, which I think was the question -- you can check me on that and let me know if I'm wrong **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 3
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17:34:18 Extra Problem (was 4.4.50) Find the equation of the line tangent to the graph of 25^(2x^2) at (-1/2,5)
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RESPONSE --> y=25^(2x^2) u=2x^2 du/dx=4x du/dx * ln(a) * a^u 4x * ln(25) * 25^(2x^2) at (-.5,5) 4(-.5) * ln 25 * 25^(2(-.5^2)) -2 ln 25 * 25^(1/2) -2 ln 25 * 5 -10 ln 25=m y-5=(-10 ln 25)(x - -.5) y-5=-10 ln 25 x -5 ln 25 y=(-10 ln 25 x) - (5 ln 25) + 5 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:39:48 Write 25^u where u = 2x^2. So du/dx = 4x. The derivative of a^x is a^x * ln(a). So the derivative of 25^u with respect to x is du / dx * ln(25) * 25^u = 4x ln(25) * 25^u = 4x ln(25) * 25^ (2 x^2). Evaluating this for x = -1/2 you get 4 * (-1/2) ln(25) * 25^(2 * (-1/2)^2 ) = -2 ln(25) * 25^(1/2) = -2 ln(25) * 5 = -10 ln(25) = -20 ln(5) = -32.189 approx. So the tangent line is a straight line thru (-1/2, 5) and having slope -20 ln(5). The equation of a straight line with slope m passing thru (x1, y1) is (y - y1) = m ( x - x1) so the slope of the tangent line must be y - 5 =-20 ln(5) ( x - (-1/2) ) or y - 5 = -20 ln(5) x - 10 ln(5). Solving for y we get y = -20 ln(5) x - 10 ln(5). A decimal approximation is y = -32.189x - 11.095 ALTERNATIVE SOLUTION: A straight line has form y - y1 = m ( x - x1), where m is the slope of the graph at the point, which is the value of the derivative of the function at the point. So you have to find the derivative of 25^(2x^2) then evaluate it at x = -1/2. The derivative of a^x is ln(a) * a^x. The derivative of 25^z would therefore be ln(25) * 25^z. The derivative of 25^(2 x^2) would be found by the chain rule with f(z) = 25^z and g(x) = 2 x^2. The result is g ' (x) * f ' (g(x)) = 4 x * ln(25) * 25^(2x^2). Evaluating at x = -1/2 we get -2 ln(25) * 25^(1/2) = -10 ln(25). Now we use the ponit-slope form of the equation of a straight line to get (y - 5) = -10 ln(25) * (x - (-1/2) ) and simplify. **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 2
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17:42:31 4.5.59 (was 4.4.59) dB = 10 log(I/10^-16); find rate of change when I=10^-4
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RESPONSE --> dB = 10 log(I/10^-16) 10 log ((10^-4)/(10^-16)) 10 log (10^12) 10*(1/ln 10)* (10^12) 4.34*10^12 confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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17:46:17 This function is a composite; the inner function is I / 10^-16, which has derivative 1/10^-16 = 10^16. So the derivative is dB' = dB / dI = 10 ( 10^16 * / ln(10) ) / (I / 10^-16) = 10 / [ ln(10) * I ]. Alternatively, 10 log(I / 10^-16) = 10 (log I - log(10^-16) ) = 10 log I + 160; the derivative comes out the same with no need of the chain rule. Plugging in I = 10^-4 we get rate = 10 / [ ln(10) * 10^-4 ] = 10^5 / ln(10), which comes out around 40,000 (use your calculator to get the accurate result. **
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RESPONSE --> I first plugged the value of I into the rate of change equation and simplified. I now see the other approaches to solving the problem would have been simpler and more accurate methods. ------------------------------------------------ Self-critique rating #$&*: 2
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18:09:10 4.5.60 (was 4.4.60) T = 87.97 + 34.96 ln p + 7.91 `sqrt(p); find rate of change
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RESPONSE --> T = 87.97 + 34.96 ln p + 7.91 `sqrt(p) dT/dp= (34.96/p) + ((1/2)*(7.91/(p^(1/2)) = (34.96/p) + (3.955/(p^(1/2)) At p=60 dT/dp=1.09 confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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18:09:19 The derivative with respect to p of ln p is 1 / p and the derivative with respect to p of sqrt(p) is 1 / (2 sqrt(p)). The derivative of the constant 89.97 is zero so dT/dp = 34.96 * 1/p + 7.91 * 1 / (2 sqrt(p)) = 34.96 / p + 3.955 / sqrt(p). **
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RESPONSE --> ------------------------------------------------ Self-critique rating #$&*: 3
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