sysD Posted August 7, 2012 Posted August 7, 2012 (edited) Here's the question I was given: Differentiate: f(x) = 3x +x3 my answer: f'(x)= (ln(3))(1)(3x) + 3x2 = (ln(3))(3x) + 3x2 However, wolfram gives me a slightly different answer: f'(x) = 3x2 + 3x(log(3)) Confirmed with two different widgets: http://www.wolframal...7197a484f4257c0 http://www.wolframal...a4fac87af8b07b2 _________________ This goes against the rule I've been taught for deriving exponential functions: f(x) = a^(g(x)) f'(x) = (ln(a))(a^g(x))(g'x) I don't really have any reference to understand why wolfram threw in the base10 log instead of the natural (base(e)) logarithm. [p.s. - i tried to throw in the math script but it returned with a bunch of garbled text and syntax errors] Edited August 7, 2012 by sysD
studiot Posted August 7, 2012 Posted August 7, 2012 Firstly you mean differentiate not derive. To derive means develop a formula for, which is what I now do [math]\begin{array}{l} y = {a^x} \\ \log y = x\log a \\ \frac{1}{y}\frac{{dy}}{{dx}} = \log a \\ \frac{{dy}}{{dx}} = {a^x}\log a \\ \end{array}[/math] Does this help? Note it does not matter what base you logs are to the formula still works
sysD Posted August 7, 2012 Author Posted August 7, 2012 Sorry, doesn't help. I don't see how the natural log and the base10 log can be used interchangably if they clearly give different answers. eg ln(3) = 1.0908612289~ log(3) = 0.477121255~
studiot Posted August 8, 2012 Posted August 8, 2012 I apologise the logarithm should only be to the base e. Some texts (including the table I looked up), and evidently Wolfram Alpha mean the natural log when they write log, something to watch out for. BTW the WA links you gave do not work for me. With that in mind you ands WA agree on the answer. 1
Daedalus Posted August 8, 2012 Posted August 8, 2012 (edited) In other words, Log[3] in Mathematica / WolframAlpha is actually [math]\text{ln} (3)[/math] where Log[10,3] is [math]\text{log}_{10} (3)[/math]: See Mathematica's help: Edited August 8, 2012 by Daedalus 1
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