What is the tangent of y=(5x)/(1+x^2),(5, 25/26) ?

The tangent of y=(5x)/(1+x^2),(5, 25/26) is y=-30/169 x+625/338

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tangent of y=(5x)/(1+x^2),(5, 25/26)

tangent of

y = \frac{5 x}{1 + x ^{2}}, (5, \frac{25}{26})

y = - \frac{30}{169} x + \frac{625}{338}

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{5 x}{1 + x ^{2}}

(5, \frac{25}{26}) : m = - \frac{30}{169}

Find the line with slope m=

- \frac{30}{169}

and passing through

(5, \frac{25}{26}) : y = - \frac{30}{169} x + \frac{625}{338}

y = - \frac{30}{169} x + \frac{625}{338}

f (x) = (\frac{x + 5}{x ^{2} + 2})^{2}

1 0^{x^{2} - 1}

\int w csc (4 w^{2} - 1) d w

x^{2} \frac{d ^{2} y}{d x ^{2}} = 0

\frac{d}{d x} (1.5 x - 0.25 x^{2})

What is the tangent of y=(5x)/(1+x^2),(5, 25/26) ?
The tangent of y=(5x)/(1+x^2),(5, 25/26) is y=-30/169 x+625/338