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Logarithm functions in basic calculus classes are defined only for positive real numbers.

But whenever we find an expression of the form $f(x)^{g(x)}$ we try to use logarithmic differentiation, we don't even care if this expression has negative values for some $x$. Why is that?

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1 Answers1

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As you said, one needs to check for if $f(x)$ is negative. However, here are two reasons for not doing so:

1) Usually $f(x)^{g(x)}$ makes sense only if $f(x) >0$. (However, this need not always be the case- for example take $g(x)=2$. But in that case you would use a different technique anyway)

2) If you want to be very formal and accurate, use the operator $T(h)=\dfrac{h'}{h}$, where $h$ is a function. This would behave exactly like the logarithmic differentiation, if you take $h(x)=f(x)^{g(x)}$ , and you can prove everything formally using this operator, instead of using logarithmic differentiation. So people tend not to worry about logarithmic differentiation, as most of the time, it's a valid method; and even when it's not entirely correct to use logarithmic differentiation, one can get the same results rigorously, using the aforementioned operator.

voldemort
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