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

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

tangent of

f (x) = \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

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

Compute the slope of

f (x) = \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{d}{d x} (3 x (18 x^{4} + \frac{13}{x + 1}))

\frac{d}{d x} (ln (sin (x^{3})))

x y = - 121

y = 12 e^{2} x

3 + 7 t + t^{2}

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