What is the tangent of f(x)=4sec(x),\at x= pi/3 ?

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

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tangent of f(x)=4sec(x),\at x= pi/3

tangent of

f (x) = 4 sec (x), at x = \frac{π}{3}

y = 83 x + 8 - \frac{8 π}{3}

Solution steps

Find the tangent point:

(\frac{π}{3}, 8)

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{π}{3} : m = 83

Find the line with slope m=

83

and passing through

(\frac{π}{3}, 8) : y = 83 x + 8 - \frac{8 π}{3}

y = 83 x + 8 - \frac{8 π}{3}

\int \frac{k}{x ^{3}} d x

\frac{d}{d x} (tan (3 x^{4}))

(sec (x) tan (x))^{^{'}}

\frac{3 ( 7 x ^{2} - 2 x + 6 )}{4 x ^{\frac{5}{2}}}

\frac{\partial}{\partial x} (x y e^{z})

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