What is the tangent of 6/(x^2+2),\at 2,1 ?

The tangent of 6/(x^2+2),\at 2,1 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),\at 2,1

tangent of

\frac{6}{x ^{2} + 2}, at 2, 1

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}}

Compute the slope of

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

x = a_{0} : 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} (cos (2 x) sin (2 x))

\int \frac{sin ( 0 ) d ^{0}}{4 + cos ^{2} ( 0 )} dd

\frac{5}{s ( s ^{2} + s + 2 )}

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

\int_{- \frac{π}{6}}^{0} 4 sec^{3} (x) d x

What is the tangent of 6/(x^2+2),\at 2,1 ?
The tangent of 6/(x^2+2),\at 2,1 is y=-(12a_{0})/((a_{0)^2+2)^2}x+(18a_{0}^2+12)/((a_{0)^2+2)^2}