What is the tangent of 1/(5+2x) ?

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

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tangent of 1/(5+2x)

tangent of

\frac{1}{5 + 2 x}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{1}{5 + 2 a _{0}})

Find the slope of

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

Compute the slope of

y = \frac{1}{5 + 2 x}

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

Find the line with slope m=

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

and passing through

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

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

f (t) = \frac{5 t + 300}{t ^{2} + 25}

\int \frac{2 e ^{2 x}}{e ^{2 x} + 8 e ^{x} + 15} d x

\int \frac{1}{x ^{2} x ^{2} + 5} d x

\int sin (x) + 9 d x

y = x, y = 5 x, y = - x + 2

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