What is the tangent of f(x)=4sec(x),\at x=(7pi)/6 ?

The tangent of f(x)=4sec(x),\at x=(7pi)/6 is y=-8/3 x-8/(sqrt(3))+(28pi)/9

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tangent of f(x)=4sec(x),\at x=(7pi)/6

tangent of

f (x) = 4 sec (x), at x = \frac{7 π}{6}

y = - \frac{8}{3} x - \frac{8}{3} + \frac{28 π}{9}

Solution steps

Find the tangent point:

(\frac{7 π}{6}, - \frac{8}{3})

Find the slope of

f (x) = 4 sec (x) : \frac{df ( x )}{d x} = 4 sec (x) tan (x)

Compute the slope of

f (x) = 4 sec (x)

x = \frac{7 π}{6} : m = - \frac{8}{3}

Find the line with slope m=

- \frac{8}{3}

and passing through

(\frac{7 π}{6}, - \frac{8}{3}) : y = - \frac{8}{3} x - \frac{8}{3} + \frac{28 π}{9}

y = - \frac{8}{3} x - \frac{8}{3} + \frac{28 π}{9}

\int \frac{2 x ^{2}}{( x - 1 ) ^{2} + 4} d x

y = (x + 2) (x - 1) (x - 3)

\int t sec^{2} (9 t) d t

\frac{d}{d x} (- 5 x^{2} + 10)

\frac{d}{d x} (\frac{1}{x ^{n}})

What is the tangent of f(x)=4sec(x),\at x=(7pi)/6 ?
The tangent of f(x)=4sec(x),\at x=(7pi)/6 is y=-8/3 x-8/(sqrt(3))+(28pi)/9