What is the tangent of f(x)=7 x/((3+x^2)),\at x=2 ?

The tangent of f(x)=7 x/((3+x^2)),\at x=2 is y=-1/7 x+16/7

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tangent of f(x)=7 x/((3+x^2)),\at x=2

tangent of

f (x) = 7 \frac{x}{( 3 + x ^{2} )}, at x = 2

y = - \frac{1}{7} x + \frac{16}{7}

Solution steps

Find the tangent point:

(2, 2)

Find the slope of

f (x) = 7 \frac{x}{3 + x ^{2}} : \frac{df ( x )}{d x} = \frac{7 ( 3 - x ^{2} )}{( 3 + x ^{2} ) ^{2}}

Compute the slope of

f (x) = 7 \frac{x}{3 + x ^{2}}

x = 2 : m = - \frac{1}{7}

Find the line with slope m=

- \frac{1}{7}

and passing through

(2, 2) : y = - \frac{1}{7} x + \frac{16}{7}

y = - \frac{1}{7} x + \frac{16}{7}

\int (y^{2} - y) d y

\frac{d}{d x} (ln (\frac{3 + e ^{2 x}}{3 - e ^{2 x}}))

\int_{0}^{N} e^{- s t} sin (3 t) d t

\frac{d}{d x} (x e^{x} + 2 e^{x})

2 y^{^{'}} = y^{3} cos (x)

What is the tangent of f(x)=7 x/((3+x^2)),\at x=2 ?
The tangent of f(x)=7 x/((3+x^2)),\at x=2 is y=-1/7 x+16/7