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

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

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tangent of f(x)= 2/(3x+1)

tangent of

f (x) = \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

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

EN : T i tl e G e n er a l Eq u a t i o n Sl o p e A t P o in t 2 Eq : 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}}

\int_{0}^{1} 7 e^{- 7 x} d x

x \to 5 lim (\frac{x + 2}{x - 4})

\int 0.1 t d t

\frac{\partial}{\partial x} (x \cdot ln (y))

\frac{d}{d x} (x^{2} + x^{2} + 1)

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