What is the tangent of y=arcsin(x/2),\at x=1 ?

The tangent of y=arcsin(x/2),\at x=1 is y= 1/(sqrt(3))x+pi/6-1/(sqrt(3))

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tangent of y=arcsin(x/2),\at x=1

tangent of

y = arcsin (\frac{x}{2}), at x = 1

y = \frac{1}{3} x + \frac{π}{6} - \frac{1}{3}

Solution steps

Find the tangent point:

(1, \frac{π}{6})

Find the slope of

y = arcsin (\frac{x}{2}) : \frac{d y}{d x} = \frac{1}{4 - x ^{2}}

Compute the slope of

y = arcsin (\frac{x}{2})

x = 1 : m = \frac{1}{3}

Find the line with slope m=

\frac{1}{3}

and passing through

(1, \frac{π}{6}) : y = \frac{1}{3} x + \frac{π}{6} - \frac{1}{3}

y = \frac{1}{3} x + \frac{π}{6} - \frac{1}{3}

\frac{d x}{d t} = sin (t) x + e^{- c o s (t)} sin (t)

x \to \infty lim (- (5))

3 y^{^{'}} + x y = 5 x, y (9) = 4

arctan (x) - arctan (x - 7)

3 e^{4 x} y^{^{'}} = - 16 \frac{x}{y ^{2}}, y (0) = 1

What is the tangent of y=arcsin(x/2),\at x=1 ?
The tangent of y=arcsin(x/2),\at x=1 is y= 1/(sqrt(3))x+pi/6-1/(sqrt(3))