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

The tangent of 6/(x^2+2) is y=-(12a_{0})/((a_{0)^2+2)^2}x+(18a_{0}^2+12)/((a_{0)^2+2)^2}

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tangent of 6/(x^2+2)

tangent of

\frac{6}{x ^{2} + 2}

y = - \frac{12 a _{0}}{( a _{0}^{2} + 2 ) ^{2}} x + \frac{18 a _{0}^{2} + 12}{( a _{0}^{2} + 2 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{6}{a _{0}^{2} + 2})

Find the slope of

y = \frac{6}{x ^{2} + 2} : \frac{d y}{d x} = - \frac{12 x}{( x ^{2} + 2 ) ^{2}}

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : m = - \frac{12 a _{0}}{( a _{0}^{2} + 2 ) ^{2}}

Find the line with slope m=

- \frac{12 a _{0}}{( a _{0}^{2} + 2 ) ^{2}}

and passing through

(a_{0}, \frac{6}{a _{0}^{2} + 2}) : y = - \frac{12 a _{0}}{( a _{0}^{2} + 2 ) ^{2}} x + \frac{18 a _{0}^{2} + 12}{( a _{0}^{2} + 2 ) ^{2}}

y = - \frac{12 a _{0}}{( a _{0}^{2} + 2 ) ^{2}} x + \frac{18 a _{0}^{2} + 12}{( a _{0}^{2} + 2 ) ^{2}}

\frac{d}{d x} (\frac{e ^{x} ( x - 2 )}{x ^{3}})

f (x) = \frac{3}{4 x + 1}

x \to 1 - lim (- \frac{1}{x - 1})

\frac{\partial}{\partial x} (x^{3} - 6 x y + 8 y^{3})

n = 1 \sum \infty \frac{n ^{0}}{4 ^{n}}

What is the tangent of 6/(x^2+2) ?
The tangent of 6/(x^2+2) is y=-(12a_{0})/((a_{0)^2+2)^2}x+(18a_{0}^2+12)/((a_{0)^2+2)^2}