What is the tangent of y=sec(x),(pi/6 ,2(sqrt(3))/3) ?

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

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tangent of y=sec(x),(pi/6 ,2(sqrt(3))/3)

tangent of

y = sec (x), (\frac{π}{6}, 2 \frac{3}{3})

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

Solution steps

Find the slope of

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

Compute the slope of

y = sec (x)

(\frac{π}{6}, 2 \frac{3}{3}) : m = \frac{2}{3}

Find the line with slope m=

\frac{2}{3}

and passing through

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

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

\frac{d y}{d x} = 1 + 8 y

tan^{8} (x)

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

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

x^{2} - 4 x y + x \frac{d y}{d x} = 0

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