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

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

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

tangent of

y = sec (x), at x = \frac{π}{3}

y = 23 x + 2 - \frac{2 π}{3}

Solution steps

Find the tangent point:

(\frac{π}{3}, 2)

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}

x \to 1 lim (- x^{2} + 4 x - 1)

f (x) = x - 5 3 x

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

10 e^{- 8 (t - 6)}

\frac{d}{d x} (\frac{x + 1}{3})

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