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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

inverse

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

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

Solution steps

Solve for

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

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

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

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

Solve for

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

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

Substitute

y = x

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

f (x) = \frac{x + 3}{2 x + 5}

(x + 2)^{3} - 5

f (x) = - \frac{1}{2} + 3

f (x) = x - (\frac{x ^{2}}{12})

f (x) = \frac{x + 1}{1 - x}

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