What is the inverse of d/d arcsec(sqrt(4x+1)) ?

The inverse of d/d arcsec(sqrt(4x+1)) is (sec^2(x)-1)/4

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inverse of d/d arcsec(sqrt(4x+1))

inverse

\frac{d}{d} arcsec (4 x + 1)

\frac{sec ^{2} ( x ) - 1}{4}

Solution steps

Solve for

arcsec (4 x + 1), y = \frac{d}{d} arcsec (4 x + 1)

arcsec (4 x + 1) = y

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

4 x + 1 = sec (y)

Solve for

x, 4 x + 1 = sec (y)

\frac{sec ^{2} ( y ) - 1}{4}

Substitute

y = x

\frac{sec ^{2} ( x ) - 1}{4}

f (x) = (26.16)

f (x) = (- 5 x + 7)^{2} + 4, x \geq \frac{7}{5}

f (x) = - x^{2} + 25, x \geq 25

f (x) = 4 - 5 x + 2

f (x) = 7^{(x)} - 5

What is the inverse of d/d arcsec(sqrt(4x+1)) ?
The inverse of d/d arcsec(sqrt(4x+1)) is (sec^2(x)-1)/4