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

The tangent of f(x)=7sec(x),\at x= pi/6 is y= 14/3 x+(14}{sqrt(3)}-\frac{7pi)/9

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

tangent of

f (x) = 7 sec (x), at x = \frac{π}{6}

y = \frac{14}{3} x + \frac{14}{3} - \frac{7 π}{9}

Solution steps

Find the tangent point:

(\frac{π}{6}, \frac{14}{3})

Find the slope of

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

Compute the slope of

f (x) = 7 sec (x)

x = \frac{π}{6} : m = \frac{14}{3}

Find the line with slope m=

\frac{14}{3}

and passing through

(\frac{π}{6}, \frac{14}{3}) : y = \frac{14}{3} x + \frac{14}{3} - \frac{7 π}{9}

y = \frac{14}{3} x + \frac{14}{3} - \frac{7 π}{9}

x y^{^{'}} + y = x

x \to a lim (\frac{1}{0})

\int_{0}^{7} x^{2} + 49 d x

\int (\frac{2}{x} + \frac{x}{2}) d x

\frac{\partial}{\partial y} (x^{2} y^{2})

What is the tangent of f(x)=7sec(x),\at x= pi/6 ?
The tangent of f(x)=7sec(x),\at x= pi/6 is y= 14/3 x+(14}{sqrt(3)}-\frac{7pi)/9