What is the tangent of f(x)=sec(x),\at x= pi/6 ?

The tangent of f(x)=sec(x),\at x= pi/6 is y= 2/3 x+2/(sqrt(3))-pi/9

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tangent of f(x)=sec(x),\at x= pi/6

tangent of

f (x) = sec (x), at x = \frac{π}{6}

y = \frac{2}{3} x + \frac{2}{3} - \frac{π}{9}

Solution steps

Find the tangent point:

(\frac{π}{6}, \frac{2 3}{3})

Find the slope of

f (x) = sec (x) : \frac{df ( x )}{d x} = sec (x) tan (x)

Compute the slope of

f (x) = sec (x)

x = \frac{π}{6} : m = \frac{2}{3}

Find the line with slope m=

\frac{2}{3}

and passing through

(\frac{π}{6}, \frac{2 3}{3}) : y = \frac{2}{3} x + \frac{2}{3} - \frac{π}{9}

y = \frac{2}{3} x + \frac{2}{3} - \frac{π}{9}

\int \frac{1}{y ^{2} + 2} d y

\frac{\partial}{\partial x} (12 - x^{2} - y^{2})

y^{^{'}} = k y (100 - y)

\frac{d}{d x} (11 sin (x))

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

What is the tangent of f(x)=sec(x),\at x= pi/6 ?
The tangent of f(x)=sec(x),\at x= pi/6 is y= 2/3 x+2/(sqrt(3))-pi/9