What is the tangent of f(x)=(-6x)/(x^2+1),\at x=1 ?

The tangent of f(x)=(-6x)/(x^2+1),\at x=1 is y=-3

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tangent of f(x)=(-6x)/(x^2+1),\at x=1

tangent of

f (x) = \frac{- 6 x}{x ^{2} + 1}, at x = 1

y = - 3

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

f (x) = \frac{- 6 x}{x ^{2} + 1} : \frac{df ( x )}{d x} = - \frac{6 ( - x ^{2} + 1 )}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

f (x) = \frac{- 6 x}{x ^{2} + 1}

x = 1 : m = 0

Find the line with slope m=

0

and passing through

(1, - 3) : y = - 3

y = - 3

\frac{\partial}{\partial x} (e^{4 x y} \cdot 4 x)

\int 4 x sec^{2} (x) d x

\frac{d}{d x} (0.2 x)

\int tan (x) sin (x) d x

\int \frac{( 4 x )}{x ^{2} + 3} d x

What is the tangent of f(x)=(-6x)/(x^2+1),\at x=1 ?
The tangent of f(x)=(-6x)/(x^2+1),\at x=1 is y=-3