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

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

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

tangent of

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

Compute the slope of

y = arccos (x) + x

x = 0 : m = 0

Find the line with slope m=

0

and passing through

(0, \frac{π}{2}) : y = \frac{π}{2}

y = \frac{π}{2}

y = 9 + 5 x^{2} - 2 x^{3}

2 y^{^{'}} + 4 y = 2 t e^{- 2 t}

\int \frac{3 - x}{3 + x} d x

\int \frac{ln ∣ x ∣}{x} d x

\int \frac{3}{x 9 - ( ln ( x ) ) ^{2}} d x

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