What is the tangent of y=(6x)/(x^2+1) ?

The tangent of y=(6x)/(x^2+1) is y=(6(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(12a_{0}^3)/((a_{0)^2+1)^2}

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tangent of y=(6x)/(x^2+1)

tangent of

y = \frac{6 x}{x ^{2} + 1}

y = \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{12 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{6 a _{0}}{a _{0}^{2} + 1})

Find the slope of

y = \frac{6 x}{x ^{2} + 1} : \frac{d y}{d x} = \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{6 x}{x ^{2} + 1}

x = a_{0} : m = \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

\frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{6 a _{0}}{a _{0}^{2} + 1}) : y = \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{12 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{6 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{12 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

\frac{\partial}{\partial x} (x sin (2 y))

f (x) = (x^{2} + 3) (x^{2} - 4 x)

\int \frac{5 x ^{2} + 7 x - 11}{x ^{2}} d x

\int (2 x + 6) e^{- 4 x} d x

\frac{\partial}{\partial x} (cos (5 x))

What is the tangent of y=(6x)/(x^2+1) ?
The tangent of y=(6x)/(x^2+1) is y=(6(-a_{0}^2+1))/((a_{0)^2+1)^2}x+(12a_{0}^3)/((a_{0)^2+1)^2}