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

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

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tangent of x/(2x+3)

tangent of

\frac{x}{2 x + 3}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = \frac{x}{2 x + 3} : \frac{d y}{d x} = \frac{3}{( 2 x + 3 ) ^{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{3}{( 2 a _{0} + 3 ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

\int_{1}^{16} \frac{6}{x} d x

\int_{2}^{5} \frac{1}{2 - 5 t} d t

9 y^{^{''}} - 6 y^{^{'}} + 4 y = 0

\frac{d}{d x} ((1 - x) e^{- x})

x \to 0 lim (\frac{4 sin ( 3 x )}{2 tan ( 5 x )})

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