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)}

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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

Compute the slope of

y = arccos (x) + x

x = a_{0} : 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}

5 x^{4}

\frac{\partial}{\partial x} (4 - x^{2} + x y - 3 y^{2})

\int 2 t sin (t) d t

\frac{\partial}{\partial x} (\frac{y}{x + y})

f (x) = 10 - x^{2}

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)}