What is the tangent of f(x)=(10x)/(1+x^2),(3,3) ?

The tangent of f(x)=(10x)/(1+x^2),(3,3) is y=-4/5 x+27/5

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tangent of f(x)=(10x)/(1+x^2),(3,3)

tangent of

f (x) = \frac{10 x}{1 + x ^{2}}, (3, 3)

y = - \frac{4}{5} x + \frac{27}{5}

Solution steps

Find the slope of

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

Compute the slope of

f (x) = \frac{10 x}{1 + x ^{2}}

(3, 3) : m = - \frac{4}{5}

Find the line with slope m=

- \frac{4}{5}

and passing through

(3, 3) : y = - \frac{4}{5} x + \frac{27}{5}

y = - \frac{4}{5} x + \frac{27}{5}

x^{5} + y^{4} x^{6} + 5 y^{^{'}} = 0

(- 5) (7.8)

x \to - \infty lim (0. 5^{x})

\int \frac{( x ^{3} + 3 x )}{x ^{2} + 1} d x

y^{^{'}} = r (1 - \frac{y}{k})

What is the tangent of f(x)=(10x)/(1+x^2),(3,3) ?
The tangent of f(x)=(10x)/(1+x^2),(3,3) is y=-4/5 x+27/5