What is the tangent of 5/((x^2+1)) ?

The tangent of 5/((x^2+1)) is y=-(10a_{0})/((a_{0)^2+1)^2}x+(15a_{0}^2+5)/((a_{0)^2+1)^2}

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tangent of 5/((x^2+1))

tangent of

\frac{5}{( x ^{2} + 1 )}

y = - \frac{10 a _{0}}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{15 a _{0}^{2} + 5}{( 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{5}{a _{0}^{2} + 1})

Find the slope of

y = \frac{5}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{10 x}{( x ^{2} + 1 ) ^{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{10 a _{0}}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

- \frac{10 a _{0}}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{5}{a _{0}^{2} + 1}) : y = - \frac{10 a _{0}}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{15 a _{0}^{2} + 5}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{10 a _{0}}{( a _{0}^{2} + 1 ) ^{2}} x + \frac{15 a _{0}^{2} + 5}{( a _{0}^{2} + 1 ) ^{2}}

\int 1 + cos (2 θ) d θ

\int \frac{x ^{2} - x - 4}{x ^{3} + 2 x} d x

\int \frac{x ^{3}}{( x + 1 ) ^{2} + 4} d x

1 6^{4} 4 x^{4} + 4

6 y^{^{'}} = x^{2}

What is the tangent of 5/((x^2+1)) ?
The tangent of 5/((x^2+1)) is y=-(10a_{0})/((a_{0)^2+1)^2}x+(15a_{0}^2+5)/((a_{0)^2+1)^2}