#$&* course Mth 272 8/8 12:57am 003. `query 3
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Given Solution: Note that ln(1-x)^(1/3) = 1/3 ln(1-x) The derivative of ln(1-x) is u ' * 1/u with u = 1-x. It follows that u ' = -1 so the derivative of ln(1-x) is -1 * 1/(1-x) = -1/(1-x). The derivative of 1/3 ln(1-x) is therefore 1/3 * -1/(1-x) = -1 / [ 3(1-x) ].** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q4.5.9 (previously 4.5.25 (was 4.4.24)) find the derivative of ln( (e^x + e^-x) / 2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: the derivative of ln(u) is 1/u du/dx; u = (e^x + e^-x)/2 so du/dx = (e^x - e^-x) / 2. The term - e^(-x) came from applying the chain rule to e^-x. The derivative of ln( (e^x + e^-x) / 2) is therefore [(e^x - e^-2) / 2 ] / ] [ (e^x + e^-x) / 2 ] = (e^x - e^-x) / (e^x + e^-x). This expression does not simplify, though it can be expressed in various forms (e.g., (1 - e^-(2x) ) / ( 1 + e^-(2x) ), obtained by dividing both numerator and denominator by e^x.). confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q 4.5.10 (previously 4.5.30 (was 4.4.30) ) write log{base 3}(x) with base e YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: 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. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: 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. ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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): ------------------------------------------------ Self-critique Rating: ********************************************* 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: 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. ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a This function is a composite; the inner function is I / 10^-16, which has derivative 1/10^-16 = 10^16. So the derivative is dB' = dB / dI = 10 ( 10^16 * / ln(10) ) / (I / 10^-16) = 10 / [ ln(10) * I ]. Alternatively, 10 log(I / 10^-16) = 10 (log I - log(10^-16) ) = 10 log I + 160; the derivative comes out the same with no need of the chain rule. Plugging in I = 10^-4 we get rate = 10 / [ ln(10) * 10^-4 ] = 10^5 / ln(10), which comes out around 40,000 (use your calculator to get the accurate result. ** STUDENT COMMENT I did not know how to find the derivative once I simplified the problem. After viewing the solution, I am still confused. INSTRUCTOR RESPONSE log I = ln(I) / ln(10). The derivative of ln(x) with respect to x is 1/x, so the derivative with respect to I of ln(I) is 1 / I. So the derivative of ln(I) / ln(10) is (1 / I ) * (1 / ln(10) ) = 1 / (I ln(10) ). &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ------------------------------------------------ Self-critique Rating: ********************************************* Question: `q4.5.26 (previously 4.5.60 (was 4.4.60)) T = 87.97 + 34.96 ln p + 7.91 `sqrt(p); find rate of change YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The derivative with respect to p of ln p is 1 / p and the derivative with respect to p of sqrt(p) is 1 / (2 sqrt(p)). The derivative of the constant 89.97 is zero so dT/dp = 34.96 * 1/p + 7.91 * 1 / (2 sqrt(p)) = 34.96 / p + 3.955 / sqrt(p). ** confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `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): ------------------------------------------------ Self-critique Rating: Note that the documents listed below are recommended as a summary, in q_a_ form, of the rules and applications of differentiation. You are welcome to submit any or all of these, or any part of any of these documents. They are recommended for anyone who isn't completely sure of the present chapter. http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa11.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa14.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa15.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: Note that the documents listed below are recommended as a summary, in q_a_ form, of the rules and applications of differentiation. You are welcome to submit any or all of these, or any part of any of these documents. They are recommended for anyone who isn't completely sure of the present chapter. http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa11.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa12.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa13.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa14.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa15.htm http://vhcc2.vhcc.edu/dsmith/genInfo/qa_query_etc/cal1/cal1_qa16.htm " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!