What is the tangent of (4x)/(x^2+1) ?

The tangent of (4x)/(x^2+1) is y=(4(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(8a_{0}^3)/((a_{0)^2+1)^2}

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

tangent of (4x)/(x^2+1)

tangent of

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

y = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{4 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = \frac{4 x}{x ^{2} + 1} : \frac{d y}{d x} = \frac{4 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

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

x = a_{0} : m = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

\frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{4 a _{0}}{a _{0}^{2} + 1}) : y = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{4 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{8 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

\int_{0}^{8} 1 + x d x

x^{3} - 5 x, 4 x, - 3, 3

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

- 3 cos (x), at x = - \frac{π}{4}

x \to \infty lim (11 + \frac{4}{x ^{2}})

What is the tangent of (4x)/(x^2+1) ?
The tangent of (4x)/(x^2+1) is y=(4(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(8a_{0}^3)/((a_{0)^2+1)^2}