What is the inverse of d/d arcsec(3x^2+1) ?

The inverse of d/d arcsec(3x^2+1) is sqrt((sec(x)-1)/3),-sqrt((sec(x)-1)/3)

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inverse of d/d arcsec(3x^2+1)

inverse

\frac{d}{d} arcsec (3 x^{2} + 1)

\frac{sec ( x ) - 1}{3}, - \frac{sec ( x ) - 1}{3}

Solution steps

Solve for

arcsec (3 x^{2} + 1), y = \frac{d}{d} arcsec (3 x^{2} + 1)

arcsec (3 x^{2} + 1) = y

arcsec (a) = b \Rightarrow a = sec (b)

3 x^{2} + 1 = sec (y)

Solve for

x, 3 x^{2} + 1 = sec (y)

\frac{sec ( y ) - 1}{3}, - \frac{sec ( y ) - 1}{3}

Substitute

y = x

\frac{sec ( x ) - 1}{3}, - \frac{sec ( x ) - 1}{3}

f (x) = 2 x^{2} + 500 x

(A^{- 1})^{T}

f (x) = \frac{2 x}{5 x - 6}

g (x) = \frac{4 x - 2}{5}

9 - x + 81

What is the inverse of d/d arcsec(3x^2+1) ?
The inverse of d/d arcsec(3x^2+1) is sqrt((sec(x)-1)/3),-sqrt((sec(x)-1)/3)