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}^{3} (1 + 6 x^{2} - 10 x^{4}) d x

(1 + p x)^{- 1}

\frac{8}{ln ( x )}

sec^{2} (x)

\int \frac{cos ( 8 x )}{cos ^{2} ( 4 x )} d x

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}