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

The tangent of ((x+3))/(2x+1) is y=-5/((2a_{0)+1)^2}x+(2a_{0}^2+12a_{0}+3)/((2a_{0)+1)^2}

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

tangent of

\frac{( x + 3 )}{2 x + 1}

y = - \frac{5}{( 2 a _{0} + 1 ) ^{2}} x + \frac{2 a _{0}^{2} + 12 a _{0} + 3}{( 2 a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

y = \frac{x + 3}{2 x + 1} : \frac{d y}{d x} = - \frac{5}{( 2 x + 1 ) ^{2}}

Compute the slope of

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

x = a_{0} : m = - \frac{5}{( 2 a _{0} + 1 ) ^{2}}

Find the line with slope m=

- \frac{5}{( 2 a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0} + 3}{2 a _{0} + 1}) : y = - \frac{5}{( 2 a _{0} + 1 ) ^{2}} x + \frac{2 a _{0}^{2} + 12 a _{0} + 3}{( 2 a _{0} + 1 ) ^{2}}

y = - \frac{5}{( 2 a _{0} + 1 ) ^{2}} x + \frac{2 a _{0}^{2} + 12 a _{0} + 3}{( 2 a _{0} + 1 ) ^{2}}

(- 8.5) (6.2)

\frac{d}{d x} ((3 x^{2}) (x^{\frac{1}{2}}))

\int_{2}^{\infty} 5 x^{- 2} d x

\int x 5 + 2 x^{2} d x

a^{2} d x = x x^{2} - a^{2} d y

What is the tangent of ((x+3))/(2x+1) ?
The tangent of ((x+3))/(2x+1) is y=-5/((2a_{0)+1)^2}x+(2a_{0}^2+12a_{0}+3)/((2a_{0)+1)^2}