#$&* course mth 272 Question: `q 4.5.5 (previously 4.5.10 (was 4.4.10)) find the derivative of ln(1-x)^(1/3)
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Given Solution: Note that ln(1-x)^(1/3) = 1/3 ln(1-x) The derivative of ln(1-x) is u ' * 1/u with u = 1-x. It follows that u ' = -1 so the derivative of ln(1-x) is -1 * 1/(1-x) = -1/(1-x). The derivative of 1/3 ln(1-x) is therefore 1/3 * -1/(1-x) = -1 / [ 3(1-x) ].** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating:3 ********************************************* Question: `q4.5.9 (previously 4.5.25 (was 4.4.24)) find the derivative of ln( (e^x + e^-x) / 2) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Ln(u)=((e^x+e^-x)/2)*((e^x-e^-x)/2) Ln(u)=(e^x+e^-x)/(e^x-e^x) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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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)=lnx/ln3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 3
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Given Solution: `a We know that log{base b}(a) = log(a) / log(b), so that log(a) = log(b) * log{base b}(a), where the 'log' in log(a) and log(b) can stand for the logarithm to any base. In particular, if this 'log' stands for the base-e logarithm, we write it as 'ln' (which stands for 'natural log'), and we could write the above as log{base b}(a) = ln(a) / ln(b). The expression in the current problem can therefore be written as log{base 3}(x) = ln(x) / ln(3). It's worth noting also that y = log{base 3}(x) means that x = 3^y; i.e., y is the power to which 3 must be raised to give us x. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: 3 ********************************************* Question: `q 4.5.10 (previously 4.5.30 (was 4.4.30) ) write log{base 3}(x) with base e YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Log{base 3}(x)=lnx/ln3 confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ 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 (if necessary): ------------------------------------------------ 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)=lnx/ln3 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 (if necessary): ------------------------------------------------ Self-critique rating: #*&!#*&!