What is the tangent of y=arccos(x)+x ?

The tangent of y=arccos(x)+x is y=(-1/(sqrt(1-a_{0)^2)}+1)x+arccos(a_{0})+(a_{0})/(sqrt(-a_{0)^2+1)}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

tangent of y=arccos(x)+x

tangent of

y = arccos (x) + x

y = (- \frac{1}{1 - a _{0}^{2}} + 1) x + arccos (a_{0}) + \frac{a _{0}}{- a _{0}^{2} + 1}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, arccos (a_{0}) + a_{0})

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 = - \frac{1}{1 - a _{0}^{2}} + 1

Find the line with slope m=

- \frac{1}{1 - a _{0}^{2}} + 1

and passing through

(a_{0}, arccos (a_{0}) + a_{0}) : y = (- \frac{1}{1 - a _{0}^{2}} + 1) x + arccos (a_{0}) + \frac{a _{0}}{- a _{0}^{2} + 1}

y = (- \frac{1}{1 - a _{0}^{2}} + 1) x + arccos (a_{0}) + \frac{a _{0}}{- a _{0}^{2} + 1}

What is the tangent of y=arccos(x)+x ?
The tangent of y=arccos(x)+x is y=(-1/(sqrt(1-a_{0)^2)}+1)x+arccos(a_{0})+(a_{0})/(sqrt(-a_{0)^2+1)}