Editing CommonMathsQuestions
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Here is a page for explanations of some of the more tricky questions that turn up occasionally. ---- !! Why is EQN:\frac{d}{dx}ln(x)=\frac{1}{x} ? To start with, it's worth looking at the graph and seeing that this is reasonable. * When /x/ is very small, /ln(x)/ is very negative and growing quickly ** so the derivative is large positive. * When /x=1,/ /ln(x)/ is zero and growing gently. ** so the derivative is about 1, although not necessarily exactly so. * As /x/ gets large, /ln(x)/ grows more slowly, but always grows ** so the derivative is positive, but getting close to 0. So it seems plausible. How about an exact calculation? We start with EQN:y=ln(x) and we want to compute EQN:\frac{dy}{dx} * EQN:y=ln(x) * => EQN:x=e^y * => EQN:\frac{d}{dy}(x)=\frac{d}{dy}(e^y) * => EQN:\frac{dx}{dy}=e^y because EQN:\frac{d}{dy}e^y=e^y * => EQN:\frac{dx}{dy}=x because EQN:x=e^y Now the really tricky part is that when /y/ is a one-to-one function of /x,/ which it is in this case if we restrict ourselves to positive /x,/ then EQN:\frac{dx}{dy}=1/\frac{dy}{dx}. To see that properly you can either use graphs and swap the co-ordinates around, or you can do the limiting process for each side. Once you accept that, we have * EQN:\frac{dy}{dx}=1/x * Hence EQN:\frac{d}{dx}ln(x)=1/x ---- !! What is 0.999999... actually equal to? This is a good one. Students seem to think that it can't be one, because it's "obviously" less than one. How do you convince them otherwise? ---- !! What is the value of EQN:0^0 ?? Hmm ... ---- !! Are there any others? ---- CategoryMaths