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

The tangent of 6/(3x+1),\at a=-1 is y=-(18)/((3a+1)^2)x+(36a+6)/((3a+1)^2)

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

tangent of

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

y = - \frac{18}{( 3 a + 1 ) ^{2}} x + \frac{36 a + 6}{( 3 a + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a

Find the tangent point:

(a, \frac{6}{3 a + 1})

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 = a : m = - \frac{18}{( 3 a + 1 ) ^{2}}

Find the line with slope m=

- \frac{18}{( 3 a + 1 ) ^{2}}

and passing through

(a, \frac{6}{3 a + 1}) : y = - \frac{18}{( 3 a + 1 ) ^{2}} x + \frac{36 a + 6}{( 3 a + 1 ) ^{2}}

y = - \frac{18}{( 3 a + 1 ) ^{2}} x + \frac{36 a + 6}{( 3 a + 1 ) ^{2}}

\int (cos (x) + \frac{1}{8} x) d x

x \to 3 - lim (e^{\frac{2}{3 - x}})

\frac{\partial}{\partial y} (x^{y} + y^{x})

\frac{\partial}{\partial x} (1 + cos (y))

x \to 0 lim (1 - cos^{3} (x))

What is the tangent of 6/(3x+1),\at a=-1 ?
The tangent of 6/(3x+1),\at a=-1 is y=-(18)/((3a+1)^2)x+(36a+6)/((3a+1)^2)