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A:

Thank you, I solved my problem. I have used qt-faststart.

Q:

Calculating the derivative of the sum of a series

Just the sum of a series,
$$\sum_{n=1}^\infty a_n$$
And the derivative of this sum,
$$\sum_{n=1}^\infty a_n – \frac{a_n}{a_{n-1} + a_{n-2}}$$
Which of the following is the correct answer?
$(a_n)(-a_n)$
$a_n(-1)$
$a_n (-a_n)$
$a_n a_{n-1}(-a_{n-1})$
I’m not sure what the $-\frac{a_n}{a_{n-1} + a_{n-2}}$ means. And what it means are the terms get multiplied by $-1$ which will cancel out the ones they are written in negative terms? Or it is the derivative of the entire thing?

A:

The derivative of a sum is not normally written like that, it is normally written like this:
$$f'(x) = \sum_{n=1}^{\infty}a_n'(x)$$
We then have
$$f'(x) = \sum_{n=1}^{\infty}a_n'(x)$$
Here the function is $f(x) = \sum_{n=1}^{\infty}a_n(x)$ and the expression is the derivative of $f$ at $x$.
So now we consider the derivative of the sum. We know that we have
$$\frac{d}{dx}\sum_{n=1}^{N}a_n(x) = a_N'(x) + \sum_{n=1}^{N-1}a_n'(x)$$
So therefore the derivative of a sum is
$$f'(x) = \sum_{n=1}^{\infty}a_n'(x)$$
and this is similar to what you say.

Q:

My app’s text tends to
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