What is the tangent of y=arcsin(x) ?

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

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tangent of y=arcsin(x)

tangent of

y = arcsin (x)

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, arcsin (a_{0}))

Find the slope of

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

Compute the slope of

y = arcsin (x)

x = a_{0} : m = \frac{1}{1 - a _{0}^{2}}

Find the line with slope m=

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

and passing through

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

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

y = (x^{2} + 3 x - 2)^{4}

\int \frac{1}{x ^{2} + 7 x + 6} d x

x \to - 2 lim (\frac{8}{x - 3})

\frac{\partial}{\partial u} (u^{9} + v^{5})

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