What is the tangent of (10x)/(1+x^2) ?

The tangent of (10x)/(1+x^2) is y=(10(1-a_{0}^2))/((1+a_{0)^2)^2}x+(20a_{0}^3)/((a_{0)^2+1)^2}

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tangent of (10x)/(1+x^2)

tangent of

\frac{10 x}{1 + x ^{2}}

y = \frac{10 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{20 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{10 a _{0}}{1 + a _{0}^{2}})

Find the slope of

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

Compute the slope of

y = \frac{10 x}{1 + x ^{2}}

x = a_{0} : m = \frac{10 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}}

Find the line with slope m=

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

and passing through

(a_{0}, \frac{10 a _{0}}{1 + a _{0}^{2}}) : y = \frac{10 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{20 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

y = \frac{10 ( 1 - a _{0}^{2} )}{( 1 + a _{0}^{2} ) ^{2}} x + \frac{20 a _{0}^{3}}{( a _{0}^{2} + 1 ) ^{2}}

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\frac{\partial}{\partial y} ((6 - x^{2} y^{2})^{2})

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What is the tangent of (10x)/(1+x^2) ?
The tangent of (10x)/(1+x^2) is y=(10(1-a_{0}^2))/((1+a_{0)^2)^2}x+(20a_{0}^3)/((a_{0)^2+1)^2}