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I am having issues with the differentiation of the following power series

$$ \large f(x) = \sum_{n \geq 1} \dfrac{(2x-2)^n}{n2^n+1}$$

I get the following result

$$ \large f'(x) = \sum_{n \geq 1} \dfrac{n(2x-2)^{n-1}}{n2^n+1}$$

but according to my professor the result is

$$ \large f'(x) = \sum_{n \geq 1} \dfrac{2n(2x-2)^{n-1}}{2n2^{n-1}+1}.$$

Can someone explain this to me or give me a hint on what I am doing wrong?

Best regards Husky

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    $\begingroup$ $\frac{d}{dx}(2x-2)^n=n(2x-2)^{n-1}2$ ... the denominator stays the same.. $\endgroup$ – nospoon Aug 15 '15 at 11:31
  • $\begingroup$ @nospoon Ahh okay, so the answer my professor provided me with is not correct? I now get it, thank you. I had similar exercises but I was completely confused when I saw that the denominator was changed. $\endgroup$ – Husky653 Aug 15 '15 at 11:36
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    $\begingroup$ @Husky653 it is correct! the denominator hasn't changed a bit! $2n2^{n-1}=n2^n$ $\endgroup$ – nospoon Aug 15 '15 at 11:36
  • $\begingroup$ the last one is actually correct since $2*2^{n-1} = 2^n$ $\endgroup$ – Husky653 Aug 15 '15 at 11:39
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    $\begingroup$ sorry I was a bit slow! haha I just realized it, I guess i just looked at (n-1) and assummed it was changed haha $\endgroup$ – Husky653 Aug 15 '15 at 11:39
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HINT: refer to the chain rule.

$$ \dfrac{d}{dx} \Big((2x -2)^n \Big) = n (2x-2)^{n-1} \dfrac{d}{dx}\Big(2x -2 \Big) = 2n(2x-2)^{n-1} $$ Therefore $$ \dfrac{df}{dx} = \dfrac{d}{dx} \left( \sum_{n \geq 1} \dfrac{(2x-2)^n}{n2^n+1}\right) = \sum_{n \geq 1}\left( \dfrac{1}{n2^n+1} \dfrac{d}{dx} \left((2x-2)^n\right)\right) = \sum_{n \geq 1}\dfrac{2n(2x-2)^{n-1}}{n2^n+1} $$

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Keep in mind that: $$\frac{\partial f(g(x))}{\partial x}=\frac{\partial f(g(x))}{\partial g(x)}\frac{\partial g(x)}{\partial x}$$

In you case $f(g(x))=(2x-2)^n$ where $g(x)=2x$, so $\dfrac{\partial g(x)}{\partial x}=2$ and $\dfrac{\partial f(g(x))}{\partial g(x)}=n(2x-2)^{n-1}$ etc...

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Here is a picture of what my professor rewrites it too

I hope this works. The text around the sum is in danish so try to ignore that, but my professor rewrites the power series and I can't really see how he removes the $2n$.

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  • $\begingroup$ @nospoon Here is a picture of the result I am provided with $\endgroup$ – Husky653 Aug 15 '15 at 12:14

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