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I know that $\frac{da^x}{dx}=a^x ln(a)$, but I tried to differentiate it in another way and I found it strange.

$$f(x)=\overbrace{0.5*0.5*0.5...}^x$$ $$ln[f(x)]=\sum_{i=1}^{x} ln(0.5)$$ $$\frac{d}{dx}ln[f(x)]=\frac{d}{dx}\sum_{i=1}^{x} ln(0.5)$$ $$\frac{1}{f(x)}\frac{df(x)}{dx}=\sum_{i=1}^{x}\frac{d}{dx} ln(0.5)$$ $$\frac{1}{f(x)}\frac{df(x)}{dx}=\sum_{i=1}^{x}0$$ $$\frac{df(x)}{dx}=0$$

Why I get a zero?

EDIT: How to do derivative with respect to the upper limit of the sum?

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    $\begingroup$ Is it true that ${d \over dx} \left(x \ln (0.5) \right) = x {d\over dx} \left( \ln(0.5) \right)$? This is the third step above. $\endgroup$ – Titus Mar 31 '17 at 18:59
  • $\begingroup$ The first equation is wrong, if $x$ isn't a non negative integer $\endgroup$ – user251257 Mar 31 '17 at 18:59
  • $\begingroup$ I don't understand why you put the sum until x. What happens if x<0? $\endgroup$ – TheWanderer Mar 31 '17 at 19:04
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    $\begingroup$ You have to derivate with respect to the upper limit of the sum. $\endgroup$ – Rafa Budría Mar 31 '17 at 19:09
  • $\begingroup$ @RafaBudría How to do derivative with respect to the upper limit of the sum? $\endgroup$ – fairytale Mar 31 '17 at 19:13
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Just differentiate the upper limit: $$\frac{1}{f(x)} \frac{df(x)}{dx} = \sum_{i=1}^{dx/dx} \ln(0.5) = \ln (0.5)$$ so that $$\frac{df(x)}{dx} = f(x) \ln (0.5) = (0.5)^x \ln (0.5).$$

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$$\frac{1}{f(x)}\frac{df(x)}{dx}=\sum_{i=1}^{x}\frac{d}{dx} ln(0.5)$$

Can't do that, you can't put the derivative inside the sum when the upper limit of the sum is a function of $x$

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