What is the tangent of y=arctan(-x),\at x=3 ?

The tangent of y=arctan(-x),\at x=3 is y=-1/10 x-arctan(3)+3/10

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tangent of y=arctan(-x),\at x=3

tangent of

y = arctan (- x), at x = 3

y = - \frac{1}{10} x - arctan (3) + \frac{3}{10}

Solution steps

Find the tangent point:

(3, - arctan (3))

Find the slope of

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

Compute the slope of

y = arctan (- x)

x = 3 : m = - \frac{1}{10}

Find the line with slope m=

- \frac{1}{10}

and passing through

(3, - arctan (3)) : y = - \frac{1}{10} x - arctan (3) + \frac{3}{10}

y = - \frac{1}{10} x - arctan (3) + \frac{3}{10}

\int 2 - x^{2} d x

\frac{d}{d x} (\frac{2}{tan ( x )})

x^{3} + 5 x^{2} - 63 x - 140

25 y^{^{''}} - 40 y^{^{'}} + 16 y = 0, y (0) = 3, y^{^{'}} (0) = - 2

\int cos (5 x) d (5 x) d x

What is the tangent of y=arctan(-x),\at x=3 ?
The tangent of y=arctan(-x),\at x=3 is y=-1/10 x-arctan(3)+3/10