What is the tangent of 2/(3x+1) ?

The tangent of 2/(3x+1) is y=-6/((3a_{0)+1)^2}x+(12a_{0}+2)/((3a_{0)+1)^2}

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tangent of 2/(3x+1)

tangent of

\frac{2}{3 x + 1}

y = - \frac{6}{( 3 a _{0} + 1 ) ^{2}} x + \frac{12 a _{0} + 2}{( 3 a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{2}{3 a _{0} + 1})

Find the slope of

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

Compute the slope of

y = \frac{2}{3 x + 1}

x = a_{0} : m = - \frac{6}{( 3 a _{0} + 1 ) ^{2}}

Find the line with slope m=

- \frac{6}{( 3 a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{2}{3 a _{0} + 1}) : y = - \frac{6}{( 3 a _{0} + 1 ) ^{2}} x + \frac{12 a _{0} + 2}{( 3 a _{0} + 1 ) ^{2}}

y = - \frac{6}{( 3 a _{0} + 1 ) ^{2}} x + \frac{12 a _{0} + 2}{( 3 a _{0} + 1 ) ^{2}}

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

\frac{d}{d x} (x^{5} e^{3 x})

\int 9 x^{5} e^{x^{6}} d x

\frac{d y}{d x}, x^{2} - x y + y^{2} = 19

\frac{\partial}{\partial x} (\frac{5 s - 3 t}{4 s + 5 t})

What is the tangent of 2/(3x+1) ?
The tangent of 2/(3x+1) is y=-6/((3a_{0)+1)^2}x+(12a_{0}+2)/((3a_{0)+1)^2}