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

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

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

tangent of

f (x) = 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

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}

y^{^{'}} + y = - e^{x}

\int (x + 1) 7 - x d x

\int_{0}^{1} t^{3} e^{- 2 t} d t

x \to 0 lim (\frac{e ^{x} - 1}{x ^{2}})

y = x^{2} - 12, y = 4 x

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