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}}

Compute the slope of

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

x = 4 : 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}

x y^{^{'}} - 2 y = x^{4}, y (1) = 0

\frac{d}{d x} ((4 x - 8) e^{4 x})

x^{5 π}

\frac{\partial}{\partial x} (10 x + 8 y)

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