What is the tangent of 4/(5x+1) ?

The tangent of 4/(5x+1) is y=-(20)/((5a_{0)+1)^2}x+(40a_{0}+4)/((5a_{0)+1)^2}

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tangent of 4/(5x+1)

tangent of

\frac{4}{5 x + 1}

y = - \frac{20}{( 5 a _{0} + 1 ) ^{2}} x + \frac{40 a _{0} + 4}{( 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{4}{5 a _{0} + 1})

Find the slope of

y = \frac{4}{5 x + 1} : \frac{d y}{d x} = - \frac{20}{( 5 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{20}{( 5 a _{0} + 1 ) ^{2}}

Find the line with slope m=

- \frac{20}{( 5 a _{0} + 1 ) ^{2}}

and passing through

(a_{0}, \frac{4}{5 a _{0} + 1}) : y = - \frac{20}{( 5 a _{0} + 1 ) ^{2}} x + \frac{40 a _{0} + 4}{( 5 a _{0} + 1 ) ^{2}}

y = - \frac{20}{( 5 a _{0} + 1 ) ^{2}} x + \frac{40 a _{0} + 4}{( 5 a _{0} + 1 ) ^{2}}

What is the tangent of 4/(5x+1) ?
The tangent of 4/(5x+1) is y=-(20)/((5a_{0)+1)^2}x+(40a_{0}+4)/((5a_{0)+1)^2}