What is the tangent of f(x)=(2x)/(1+x^2),\at x=4 ?

The tangent of f(x)=(2x)/(1+x^2),\at x=4 is y=-30/289 x+256/289

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tangent of f(x)=(2x)/(1+x^2),\at x=4

tangent of

f (x) = \frac{2 x}{1 + x ^{2}}, at x = 4

y = - \frac{30}{289} x + \frac{256}{289}

Solution steps

Find the tangent point:

(4, \frac{8}{17})

Find the slope of

f (x) = \frac{2 x}{1 + x ^{2}} : \frac{df ( x )}{d x} = \frac{2 ( 1 - x ^{2} )}{( 1 + x ^{2} ) ^{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{30}{289}

Find the line with slope m=

- \frac{30}{289}

and passing through

(4, \frac{8}{17}) : y = - \frac{30}{289} x + \frac{256}{289}

y = - \frac{30}{289} x + \frac{256}{289}

y^{^{''}} - 3 y^{^{'}} + 2 y = e^{x} + 2 e^{x} + e^{- x}

\frac{\partial}{\partial x} (x^{3})

\int 2 + sin (x) d x

\int_{0}^{1} \frac{1}{x ^{1.2}} d x

- \frac{e ^{\frac{1}{x}}}{x ^{2}}

What is the tangent of f(x)=(2x)/(1+x^2),\at x=4 ?
The tangent of f(x)=(2x)/(1+x^2),\at x=4 is y=-30/289 x+256/289