What is the tangent of y=7sec(x),\at x= pi/3 ?

The tangent of y=7sec(x),\at x= pi/3 is y=14sqrt(3)x+14-(14pi)/(sqrt(3))

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tangent of y=7sec(x),\at x= pi/3

tangent of

y = 7 sec (x), at x = \frac{π}{3}

y = 143 x + 14 - \frac{14 π}{3}

Solution steps

Find the tangent point:

(\frac{π}{3}, 14)

Find the slope of

y = 7 sec (x) : \frac{d y}{d x} = 7 sec (x) tan (x)

Compute the slope of

y = 7 sec (x)

x = \frac{π}{3} : m = 143

Find the line with slope m=

143

and passing through

(\frac{π}{3}, 14) : y = 143 x + 14 - \frac{14 π}{3}

y = 143 x + 14 - \frac{14 π}{3}

\frac{d}{d t} (arctan (\frac{t}{2}))

x^{2} + 5

y = \frac{sec ( x )}{1 + sec ( x )}

\int 19 sin^{- \frac{3}{2}} (x) cos^{3} (x) d x

f (x) = \frac{4 - x ^{2}}{8 + x ^{2}}

What is the tangent of y=7sec(x),\at x= pi/3 ?
The tangent of y=7sec(x),\at x= pi/3 is y=14sqrt(3)x+14-(14pi)/(sqrt(3))