What is the tangent of 1/(3x^2+1) ?

The tangent of 1/(3x^2+1) is y=-(6a_{0})/((3a_{0)^2+1)^2}x+(9a_{0}^2+1)/((3a_{0)^2+1)^2}

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tangent of 1/(3x^2+1)

tangent of

\frac{1}{3 x ^{2} + 1}

y = - \frac{6 a _{0}}{( 3 a _{0}^{2} + 1 ) ^{2}} x + \frac{9 a _{0}^{2} + 1}{( 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{1}{3 a _{0}^{2} + 1})

Find the slope of

y = \frac{1}{3 x ^{2} + 1} : \frac{d y}{d x} = - \frac{6 x}{( 3 x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{1}{3 x ^{2} + 1}

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

Find the line with slope m=

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

and passing through

(a_{0}, \frac{1}{3 a _{0}^{2} + 1}) : y = - \frac{6 a _{0}}{( 3 a _{0}^{2} + 1 ) ^{2}} x + \frac{9 a _{0}^{2} + 1}{( 3 a _{0}^{2} + 1 ) ^{2}}

y = - \frac{6 a _{0}}{( 3 a _{0}^{2} + 1 ) ^{2}} x + \frac{9 a _{0}^{2} + 1}{( 3 a _{0}^{2} + 1 ) ^{2}}

n = 1 \sum \infty n^{- \frac{2}{3}}

2 (t - 3)^{2} + 9 (t - 3) + 11

y = 3 e^{4 x} + 6 e^{2 x}

\frac{d}{d x} (e^{7 x} + e^{- 7 x})

\frac{d}{d x} (1 + (\frac{3}{2} x^{(\frac{1}{2})})^{2})

What is the tangent of 1/(3x^2+1) ?
The tangent of 1/(3x^2+1) is y=-(6a_{0})/((3a_{0)^2+1)^2}x+(9a_{0}^2+1)/((3a_{0)^2+1)^2}