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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Imagine that you need to compute e^0.4 but you have no calculator or other aid to enable you to compute it exactly, only paper a

labwork [276]

Answer:

0.0032

Step-by-step explanation:

We need to compute $e^{0.4}$ by the help of third-degree Taylor polynomial that is expanded around at x = 0.

Given :

$e^{0.4}$ < e < 3

Therefore, the Taylor's Error Bound formula is given by :

$|\text{Error}| \leq \frac{M}{(N+1)!} |x-a|^{N+1}$ , where $M=|F^{N+1}(x)|$

$\leq \frac{3}{(3+1)!} |-0.4|^4$

$\leq \frac{3}{24} \times (0.4)^4$

$\leq 0.0032$

Therefore, |Error| ≤ 0.0032

4 0

3 years ago

Will mark brainliest! Questions are in the picture!

alexdok [17]

Answer:

52.9 inches .

Step-by-step explanation:

Given that the triangular indentation has an area of 100 in.² and the base and height of this traingle are represented by expressions 3x and x+3 respectively . We need to find out the perimeter to the nearest tenth.

As we know that the area of triangle is ,

$\rm\longrightarrow Area_{\triangle}=\dfrac{1}{2}(base)(height)$

Substituting the respective values,

$\rm\longrightarrow 100\ =\dfrac{1}{2}(x+3)(3x)\\\\\rm\longrightarrow 2(100) = 3x(x+3)\\\\\rm\longrightarrow200 = 3x^2+9x\\\\\rm\longrightarrow 3x^2+9x-200=0$

On using the Quadratic formula , we have;

$\rm\longrightarrow x =\dfrac{-9\pm\sqrt{9^2-4(3)(-200)}}{2(3)} \\\\\rm\longrightarrow x =\dfrac{-9\pm \sqrt{81+1200}}{6}\\\\\rm\longrightarrow x =\dfrac{-9\pm \sqrt{1281}}{6}$