What is the tangent of y=arccos(x)+x,\at x=0 ?

The tangent of y=arccos(x)+x,\at x=0 is y= pi/2

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tangent of y=arccos(x)+x,\at x=0

tangent of

y = arccos (x) + x, at x = 0

y = \frac{π}{2}

Solution steps

Find the tangent point:

(0, \frac{π}{2})

Find the slope of

y = arccos (x) + x : \frac{d y}{d x} = - \frac{1}{1 - x ^{2}} + 1

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 = 0

Find the line with slope m=

0

and passing through

(0, \frac{π}{2}) : y = \frac{π}{2}

y = \frac{π}{2}

t \frac{d x}{d t} = x - 2 t

n = 1 \sum \infty 2 + \frac{n}{3 n + 2}

\frac{s ^{2}}{( s + 5 ) ( s ^{2} + 2 ^{2} )}

\int \frac{( ln ( x ) ) ^{3}}{x} d x

\int sin (2 x - \frac{π}{4}) d x

What is the tangent of y=arccos(x)+x,\at x=0 ?
The tangent of y=arccos(x)+x,\at x=0 is y= pi/2