What is the tangent of (8x)/((x^2+1)) ?

The tangent of (8x)/((x^2+1)) is y=(8(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(16a_{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 (8x)/((x^2+1))

tangent of

\frac{8 x}{( x ^{2} + 1 )}

y = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{16 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{8 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = \frac{8 x}{x ^{2} + 1} : \frac{d y}{d x} = \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{8 x}{x ^{2} + 1}

x = a_{0} : m = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

\frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{8 a _{0}}{a _{0}^{2} + 1}) : y = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{16 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{16 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

x \to 0 - lim (ln (x))

\int \frac{8}{( 4 - t ) ^{8}} d t

y = 8 + 4 x^{2} - 2 x^{3}, (2, 8)

x \to 2 lim (x^{2} + 3 x - 4)

\int \frac{x}{x ^{4} + 7} d x

What is the tangent of (8x)/((x^2+1)) ?
The tangent of (8x)/((x^2+1)) is y=(8(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(16a_{0}^3)/((a_{0)^2+1)^2}