3 years ago

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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In the diagram of circle O, what is the measure of ? 27° 54° 108° 120°

vesna_86 [32]

Answer:

Option (2).

Step-by-step explanation:

This question is incomplete; here is the complete question and find the figure attached.

In the diagram of a circle O, what is the measure of ∠ABC?

27°

54°

108°

120°

m(minor arc $\widehat {AC}$ ) = 126°

m(major arc $\widehat {AC}$ ) = 234°

By the intersecting tangents theorem,

If the two tangents of a circle intersect each other outside the circle, measure of angle formed between them is half the difference of the measures of the intercepted arcs.

m∠ABC = $\frac{1}{2}[m(\text {major arc{AC})}-m(\text{minor arc} {AC})]$