#$&* course Mth 272 6/17 12:47 am Question: `q 4.5.5 (previously 4.5.10 (was 4.4.10)) find the derivative of ln(1-x)^(1/3)
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Given Solution: Note that ln(1-x)^(1/3) = 1/3 ln(1-x) The derivative of ln(1-x) is u ' * 1/u with u = 1-x. It follows that u ' = -1 so the derivative of ln(1-x) is -1 * 1/(1-x) = -1/(1-x). The derivative of 1/3 ln(1-x) is therefore 1/3 * -1/(1-x) = -1 / [ 3(1-x) ].** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q4.5.9 (previously 4.5.25 (was 4.4.24)) find the derivative of ln( (e^x + e^-x) / 2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ln((e^x + e^-x) / 2) ((e^x + e^-x) / 2) / ((e^x - e^-x) / 2) (e^x + e^-x) / (e^x - e^-x) confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q 4.5.10 (previously 4.5.30 (was 4.4.30) ) write log{base 3}(x) with base e YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: log{base 3}(x) = ln(x) / ln(3) since the “log” is of base e and e and ln are inverses. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a We know that log{base b}(a) = log(a) / log(b), so that log(a) = log(b) * log{base b}(a), where the 'log' in log(a) and log(b) can stand for the logarithm to any base. In particular, if this 'log' stands for the base-e logarithm, we write it as 'ln' (which stands for 'natural log'), and we could write the above as log{base b}(a) = ln(a) / ln(b). The expression in the current problem can therefore be written as log{base 3}(x) = ln(x) / ln(3). It's worth noting also that y = log{base 3}(x) means that x = 3^y; i.e., y is the power to which 3 must be raised to give us x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q4.5.22 (previously extra prob (was 4.4.50)). Find the equation of the line tangent to the graph of 25^(2x^2) at (-1/2,5) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: u = 2x^2 du/dx = 2(2x) = 4x 25^u = ln(25) 4x * ln (25) * 25^(2x^2) = 4(-.5) * ln(25) * 25^(2(-.5)^2) = -20 ln(5) = -32.189 y - 5 = -32.189(x - 1/2) y = -32.189x - 16.095 + 5 y = -32.189 - 11.095 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I created the equation without having to revert to the ln notation. ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `q4.5.25 (previously 4.5.59 (was 4.4.59)) dB = 10 log(I/10^-16); find rate of change when I=10^-4 YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: (1 I/10^-16) = 1/10^-16 = 10^16 = (10^16 ln(10)) / (I / 10 ^-16) = 10 / (ln(10) * I) 10 / (ln(10) * 10^-4) = 43,429.45 confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a 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. ** STUDENT COMMENT I did not know how to find the derivative once I simplified the problem. After viewing the solution, I am still confused. INSTRUCTOR RESPONSE log I = ln(I) / ln(10). The derivative of ln(x) with respect to x is 1/x, so the derivative with respect to I of ln(I) is 1 / I. So the derivative of ln(I) / ln(10) is (1 / I ) * (1 / ln(10) ) = 1 / (I ln(10) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Would that number be considered close? ------------------------------------------------ Self-critique Rating: OK
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Given Solution: 3 `a 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). ** STUDENT QUESTION Not understanding why the sqrt does not go away INSTRUCTOR RESPONSE The derivative with respect to p of sqrt(p) is 1 / (2 sqrt(p)). This is a familiar derivative from first-semester calculus, obtained from the power-function rule that tells us that the derivative of x^a is a x^(a - 1). (this is usually stated with exponent p instead of a, but since p is the variable in this problem that form would almost certainly be confusing). For example the derivative of x^3 is 3 * x^(3-1) = 3 x^2. The derivative relevant to the current problem is the derivative of sqrt(x), or x^(1/2). The derivative is 1/2 * x^(1/2 - 1) = 1/2 x ^ (-1/2). x^(-1/2) = 1 / x^(1/2) = 1 / sqrt(x). Therefore our derivative 1/2 x^(-1/2) is 1 / (2 sqrt(x)). In the current problem the variable is p rather than x. The derivative, with respect to p, of p^(1/2) is 1/2 p^(1/2 - 1) = 1 / (2 sqrt(p) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!