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}}

Compute the slope of

y = \frac{4}{5 x + 1}

x = a_{0} : 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}}

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

\frac{d y}{d t} = \frac{y}{t + 1} + 4 t^{2} + 4 t, y (1) = - 5

x \to 0 lim (\frac{1}{x} - 3)

f (y) = \frac{y - 1}{y ^{2} - y + 1}

\int (x + 1) 4 x + 5 d x

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}