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

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

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

tangent of

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

y = - 3

Solution steps

Find the tangent point:

(1, - 3)

Find the slope of

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

Compute the slope of

y = - \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

f (x) = 9 x^{2}

\int \frac{x}{1 - 5 x ^{2}} d x

\frac{d}{d x} (\frac{1}{x ^{13}})

\int 4 tan^{3} (x) sec (x) d x

\int_{- 1}^{0} (x + 1)^{3} d x

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