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

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

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

tangent of

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}

- x^{2} + 3 x, 2 x^{3} - x^{2} - 5 x

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

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

(1, 7), (7, 3)

\frac{d}{d θ} (4 sin^{2} (θ) cos (θ))

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