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

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

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tangent of y= 5/(4x+1)

tangent of

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

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{5}{4 a _{0} + 1})

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

- \frac{20}{( 4 a _{0} + 1 ) ^{2}}

and passing through

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

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

\frac{d}{d x} (\frac{3 - 17 x ^{2}}{3 x ^{3}})

5 e^{x} tan (x)

\int \frac{4 x ^{2} + 3 x - 1}{x ^{3} - x ^{2}} d x

\frac{\partial}{\partial y} (\frac{y}{x + z})

\int \frac{2 x + 1}{x + 1} d x

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