What is the derivative of arcsec(3x^2+1) ?

The derivative of arcsec(3x^2+1) is (2sqrt(3))/((3x^2+1)sqrt(3x^2+2))

What is the first derivative of arcsec(3x^2+1) ?

The first derivative of arcsec(3x^2+1) is (2sqrt(3))/((3x^2+1)sqrt(3x^2+2))

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

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

Solution steps

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

Apply the chain rule:

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

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

\frac{d}{d x} (3 x^{2} + 1) = 6 x

= \frac{1}{( 3 x ^{2} + 1 ) ^{2} ( 3 x ^{2} + 1 ) ^{2} - 1} \cdot 6 x

Simplify

\frac{1}{( 3 x ^{2} + 1 ) ^{2} ( 3 x ^{2} + 1 ) ^{2} - 1} \cdot 6 x : \frac{2 3}{( 3 x ^{2} + 1 ) 3 x ^{2} + 2}

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

x \to \infty lim (n c^{n})

x \to 0 lim (\frac{3 x}{x + 5})

y^{^{'}} + y = e^{6 t}, y (0) = - 1

y^{^{'}} = sin (5 x)

h^{^{'}} = - h^{\frac{1}{3}}

What is the derivative of arcsec(3x^2+1) ?
The derivative of arcsec(3x^2+1) is (2sqrt(3))/((3x^2+1)sqrt(3x^2+2))
What is the first derivative of arcsec(3x^2+1) ?
The first derivative of arcsec(3x^2+1) is (2sqrt(3))/((3x^2+1)sqrt(3x^2+2))