What is the tangent of f(x)=tan^2(x),\at x=(-pi)/4 ?

The tangent of f(x)=tan^2(x),\at x=(-pi)/4 is y=-4x+1-pi

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of f(x)=tan^2(x),\at x=(-pi)/4

tangent of

f (x) = tan^{2} (x), at x = \frac{- π}{4}

y = - 4 x + 1 - π

Solution steps

Find the tangent point:

(\frac{- π}{4}, 1)

Find the slope of

f (x) = tan^{2} (x) : \frac{df ( x )}{d x} = 2 tan (x) sec^{2} (x)

Compute the slope of

f (x) = tan^{2} (x)

x = \frac{- π}{4} : m = - 4

Find the line with slope m=

- 4

and passing through

(\frac{- π}{4}, 1) : y = - 4 x + 1 - π

y = - 4 x + 1 - π

\frac{\partial}{\partial x} ((2 x - 5 y)^{2})

\int y^{3} + 10 x y^{4} - 2 x d x

x \to - 4 - lim (\frac{x + 3}{x + 4})

f (x) = x 8 - x^{2}

\frac{d}{d x} (\frac{x + 2}{x + 3})

What is the tangent of f(x)=tan^2(x),\at x=(-pi)/4 ?
The tangent of f(x)=tan^2(x),\at x=(-pi)/4 is y=-4x+1-pi