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

The derivative of y=sqrt(x^2-1)arcsec(x) is (xarcsec(x))/(sqrt(x^2-1))+1/(sqrt(x^2))

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

The first derivative of y=sqrt(x^2-1)arcsec(x) is (xarcsec(x))/(sqrt(x^2-1))+1/(sqrt(x^2))

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derivative of

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

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

Solution steps

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2} - 1, g = arcsec (x)

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

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

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

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

Simplify

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

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

6 (t) e^{- 2 t}

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

y = \frac{t}{( t - 5 ) ^{2}}

\int 5 cos^{3} (6 x) d x

\frac{\partial}{\partial x} (3 x^{3} + x^{2} y - 3 y^{3})

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