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

The tangent of y=(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}

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tangent of y=(8x)/((x^2+1))

tangent of

y = \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}}

\int e^{l n (x^{5})} d x

(2 x - 3)^{4}

\frac{d}{d x} (e^{x} ln (x^{5}))

x^{^{'}} + x = e^{t}

\int_{0}^{15} x 15 - x d x

What is the tangent of y=(8x)/((x^2+1)) ?
The tangent of y=(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}