What is the tangent of f(x)=sec(x),(pi/3 ,2) ?

The tangent of f(x)=sec(x),(pi/3 ,2) is y=2sqrt(3)x+2-(2pi)/(sqrt(3))

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=sec(x),(pi/3 ,2)

tangent of

f (x) = sec (x), (\frac{π}{3}, 2)

y = 23 x + 2 - \frac{2 π}{3}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = sec (x)

(\frac{π}{3}, 2) : m = 23

Find the line with slope m=

23

and passing through

(\frac{π}{3}, 2) : y = 23 x + 2 - \frac{2 π}{3}

y = 23 x + 2 - \frac{2 π}{3}

\frac{\partial}{\partial y} (sinh (2 x + 3 y))

\frac{55.9}{t ^{2} - 0.31 t + 11.2}

\frac{\partial}{\partial x} (- 3 x^{2} + 28)

\frac{d P}{d t} = P (5 - P) - 4

x \to 2 lim (x^{3} \cdot 2 x^{2})

What is the tangent of f(x)=sec(x),(pi/3 ,2) ?
The tangent of f(x)=sec(x),(pi/3 ,2) is y=2sqrt(3)x+2-(2pi)/(sqrt(3))