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

The tangent of 6/(3x+1),\at x=-1 is y=-9/2 x-15/2

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tangent of 6/(3x+1),\at x=-1

tangent of

\frac{6}{3 x + 1}, at x = - 1

y = - \frac{9}{2} x - \frac{15}{2}

Solution steps

Find the tangent point:

(- 1, - 3)

Find the slope of

y = \frac{6}{3 x + 1} : \frac{d y}{d x} = - \frac{18}{( 3 x + 1 ) ^{2}}

Compute the slope of

y = \frac{6}{3 x + 1}

x = - 1 : m = - \frac{9}{2}

Find the line with slope m=

- \frac{9}{2}

and passing through

(- 1, - 3) : y = - \frac{9}{2} x - \frac{15}{2}

y = - \frac{9}{2} x - \frac{15}{2}

\frac{\partial}{\partial x} (- x - 6)

\frac{\partial}{\partial x} (4 - x^{2} - y^{2})

121 y^{^{''}} - 154 y^{^{'}} + 85 y = 0

- \frac{4}{3} x^{\frac{2}{3}}

\frac{\partial}{\partial x} (e^{- x} cos (yz) z)

What is the tangent of 6/(3x+1),\at x=-1 ?
The tangent of 6/(3x+1),\at x=-1 is y=-9/2 x-15/2