What is the tangent of y= 5/(1+x^2),(4,0.29412) ?

The tangent of y= 5/(1+x^2),(4,0.29412) is y=-40/289 x+0.29412+160/289

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tangent of y= 5/(1+x^2),(4,0.29412)

tangent of

y = \frac{5}{1 + x ^{2}}, (4, 0.29412)

y = - \frac{40}{289} x + 0.29412 + \frac{160}{289}

Solution steps

Find the slope of

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

Compute the slope of

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

(4, 0.29412) : m = - \frac{40}{289}

Find the line with slope m=

- \frac{40}{289}

and passing through

(4, 0.29412) : y = - \frac{40}{289} x + 0.29412 + \frac{160}{289}

y = - \frac{40}{289} x + 0.29412 + \frac{160}{289}

f (x) = 6 x^{2} cot (x)

x \to \infty lim (2 x - 2 x)

\frac{d}{d x} (x^{8} cos (x))

\frac{\partial}{\partial y} (y^{4} + 5 x y^{3})

\int (54) d x

What is the tangent of y= 5/(1+x^2),(4,0.29412) ?
The tangent of y= 5/(1+x^2),(4,0.29412) is y=-40/289 x+0.29412+160/289