What is the tangent of f(x)=(-8x)/(x^2+1) ?

The tangent of f(x)=(-8x)/(x^2+1) is y=-(8(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(16a_{0}^3)/((a_{0)^2+1)^2}

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tangent of f(x)=(-8x)/(x^2+1)

tangent of

f (x) = \frac{- 8 x}{x ^{2} + 1}

y = - \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{16 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{8 a _{0}}{a _{0}^{2} + 1})

Find the slope of

f (x) = \frac{- 8 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{8 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{- 8 x}{x ^{2} + 1}

x = a_{0} : m = - \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

Find the line with slope m=

- \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}}

and passing through

(a_{0}, - \frac{8 a _{0}}{a _{0}^{2} + 1}) : y = - \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{16 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = - \frac{8 ( - a _{0}^{2} + 1 )}{( a _{0}^{2} + 1 ) ^{2}} x - \frac{16 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

3 t + 2

x \to - 2 lim (x^{2} + 2 x - 3)

4 cos (x)

2 x^{2} y^{^{'}} = y^{^{'}} + 5 x e^{- y}

\int \frac{4}{sec ( x ) e ^{s i n (x)}} d x

What is the tangent of f(x)=(-8x)/(x^2+1) ?
The tangent of f(x)=(-8x)/(x^2+1) is y=-(8(-a_{0}^2+1))/((a_{0)^2+1)^2}x-(16a_{0}^3)/((a_{0)^2+1)^2}