phinah 29 Aug 2017, 19:07
In the chapter Derivative of the Logarithmic Function, Example #6, is it necessary to apply the change of base to get to the right solution?
5. Derivative of the Logarithmic Function
Based upon the fact that
`dy/dx ` `= 1/["argument" xx ln(base)] ` `xx [d/dx("argument")]`
my solution was
`dy/dx = 1/[6x ln 2] * (6) = 1/[x ln 2]`
X
In the chapter Derivative of the Logarithmic Function, Example #6, is it necessary to apply the change of base to get to the right solution?
Relevant page
<a href="/differentiation-transcendental/5-derivative-logarithm.php">5. Derivative of the Logarithmic Function</a>
What I've done so far
Based upon the fact that
`dy/dx ` `= 1/["argument" xx ln(base)] ` `xx [d/dx("argument")]`
my solution was
`dy/dx = 1/[6x ln 2] * (6) = 1/[x ln 2]`Murray 30 Aug 2017, 21:54
@Phinah
Please use the math entry system so I, and others, can read your question. I have edited it just now.
I checked your answer by finding `1/(ln(2))` and it has the same value, `1.4427`, so your approach appears to be fine!
Whenever I see a log expression with a base other than `e`, I automatically change base. This means I only need to learn one formula and can apply it in many places. In this case, I like the look of your approach better! :-)
X
@Phinah Please use the math entry system so I, and others, can read your question. I have edited it just now. I checked your answer by finding `1/(ln(2))` and it has the same value, `1.4427`, so your approach appears to be fine! Whenever I see a log expression with a base other than `e`, I automatically change base. This means I only need to learn one formula and can apply it in many places. In this case, I like the look of your approach better! :-)
X
Ok got it! Thank you for the explanation.
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