What is the tangent of f(x)=(-6x)/(x^2+1),\at x=0 ?

The tangent of f(x)=(-6x)/(x^2+1),\at x=0 is y=-6x

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tangent of f(x)=(-6x)/(x^2+1),\at x=0

tangent of

f (x) = \frac{- 6 x}{x ^{2} + 1}, at x = 0

y = - 6 x

Solution steps

Find the tangent point:

(0, 0)

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 = 0 : m = - 6

Find the line with slope m=

- 6

and passing through

(0, 0) : y = - 6 x

y = - 6 x

\int tan (θ) sec (θ) sec^{2} (θ) d θ

\frac{d}{d x} (ln (4 + ln (x)))

\frac{\partial}{\partial x} (a x ln (a x))

\int 49 y d y

e^{- 4 t}

What is the tangent of f(x)=(-6x)/(x^2+1),\at x=0 ?
The tangent of f(x)=(-6x)/(x^2+1),\at x=0 is y=-6x