What is the tangent of y=5sec(x),\at x= pi/3 ?

The tangent of y=5sec(x),\at x= pi/3 is y=10sqrt(3)x+10-(10pi)/(sqrt(3))

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tangent of y=5sec(x),\at x= pi/3

tangent of

y = 5 sec (x), at x = \frac{π}{3}

y = 103 x + 10 - \frac{10 π}{3}

Solution steps

Find the tangent point:

(\frac{π}{3}, 10)

Find the slope of

y = 5 sec (x) : \frac{d y}{d x} = 5 sec (x) tan (x)

Compute the slope of

y = 5 sec (x)

x = \frac{π}{3} : m = 103

Find the line with slope m=

103

and passing through

(\frac{π}{3}, 10) : y = 103 x + 10 - \frac{10 π}{3}

y = 103 x + 10 - \frac{10 π}{3}

\frac{d}{d x} (- 5 e^{- 4 x})

\int_{0}^{2} (x + 1) d x

y = 1 + \frac{2}{x ^{4}}

\int y + y (3 + y^{2})^{\frac{1}{2}} d y

\int x^{3} 100 - x^{2} d x

What is the tangent of y=5sec(x),\at x= pi/3 ?
The tangent of y=5sec(x),\at x= pi/3 is y=10sqrt(3)x+10-(10pi)/(sqrt(3))