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

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

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

tangent of

y = sec (x), (\frac{π}{3}, 2)

y = 23 x + 2 - \frac{2 π}{3}

Solution steps

Find the slope of

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = 23

Find the line with slope m=

23

and passing through

(\frac{π}{3}, 2) : y = 23 x + 2 - \frac{2 π}{3}

y = 23 x + 2 - \frac{2 π}{3}

\frac{7}{6 s ( s + \frac{2}{9} )}

\int ln (x + 9) d x

\int e^{2} d x

\frac{d}{d x} ((9 x + 8) x^{2})

y^{^{'}} = (1 - 2 x) y^{2}, y (0) = - \frac{1}{6}

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