What is the tangent of ((3x+5))/(1+x) ?

The tangent of ((3x+5))/(1+x) is y=-2/((1+a_{0))^2}x+(3a_{0}^2+10a_{0}+5)/((a_{0)+1)^2}

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tangent of ((3x+5))/(1+x)

tangent of

\frac{( 3 x + 5 )}{1 + x}

y = - \frac{2}{( 1 + a _{0} ) ^{2}} x + \frac{3 a _{0}^{2} + 10 a _{0} + 5}{( a _{0} + 1 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{3 a _{0} + 5}{1 + a _{0}})

Find the slope of

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

Compute the slope of

y = \frac{3 x + 5}{1 + x}

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

Find the line with slope m=

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

and passing through

(a_{0}, \frac{3 a _{0} + 5}{1 + a _{0}}) : y = - \frac{2}{( 1 + a _{0} ) ^{2}} x + \frac{3 a _{0}^{2} + 10 a _{0} + 5}{( a _{0} + 1 ) ^{2}}

y = - \frac{2}{( 1 + a _{0} ) ^{2}} x + \frac{3 a _{0}^{2} + 10 a _{0} + 5}{( a _{0} + 1 ) ^{2}}

ln (1 + 2 x)

\int ln^{3} (2 x) d x

\frac{1}{x - 3}

\frac{d}{d x} (ln (y x))

x y^{^{'}} - (1 + x) y = x y^{2}

What is the tangent of ((3x+5))/(1+x) ?
The tangent of ((3x+5))/(1+x) is y=-2/((1+a_{0))^2}x+(3a_{0}^2+10a_{0}+5)/((a_{0)+1)^2}