How find derivative of function: y=arcsin(2x+1) ...?

1 answer:

3 0

$( \arcsin{x})'= \frac{1}{ \sqrt{1-x^2} } 
\\
\\ (\arcsin{(2x+1)})'= \frac{1}{ \sqrt{1-(2x+1)^2} } (2x+1)'= \frac{2}{ \sqrt{1-(4x^2+4x+1)} } =
\\
\\= \frac{2}{ \sqrt{-4x^2-4x} } =\frac{2}{ \sqrt{4x(-x-1} )} =\frac{2}{ 2\sqrt{x(-x-1} )} =\frac{1}{ \sqrt{x(-x-1} )}$

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1. Given the below sequence: -1, -3, -5, -7, . . . (a) What are the next 3 terms? (b) Is this an arithmetic or geometric sequenc

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(a) To find the next three terms , we must firs check if it is arithmetic sequence or a geometric sequence . For it to be an arithmetic sequence , there must be a common difference :

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