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

The derivative of y=arcsec(x^2+1) is (2x)/(sqrt(x^2)\sqrt{(x^2+1)^2)sqrt(x^2+2)}

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

The first derivative of y=arcsec(x^2+1) is (2x)/(sqrt(x^2)\sqrt{(x^2+1)^2)sqrt(x^2+2)}

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derivative of

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

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

Solution steps

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

Apply the chain rule:

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

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

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

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

Simplify

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

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

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

x \to \infty lim (\frac{2}{x} + 1)

\int 10 sin (t) + 3 cos (t) d t

\frac{\partial}{\partial x} (x y)

\frac{d}{d x} (\frac{e ^{x}}{1 + x})

What is the derivative of y=arcsec(x^2+1) ?
The derivative of y=arcsec(x^2+1) is (2x)/(sqrt(x^2)\sqrt{(x^2+1)^2)sqrt(x^2+2)}
What is the first derivative of y=arcsec(x^2+1) ?
The first derivative of y=arcsec(x^2+1) is (2x)/(sqrt(x^2)\sqrt{(x^2+1)^2)sqrt(x^2+2)}