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

The tangent of 1/(5+3x) is y=-3/((5+3a_{0))^2}x+(6a_{0}+5)/((3a_{0)+5)^2}

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tangent of 1/(5+3x)

tangent of

\frac{1}{5 + 3 x}

y = - \frac{3}{( 5 + 3 a _{0} ) ^{2}} x + \frac{6 a _{0} + 5}{( 3 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 + 3 a _{0}})

Find the slope of

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

Find the line with slope m=

- \frac{3}{( 5 + 3 a _{0} ) ^{2}}

and passing through

(a_{0}, \frac{1}{5 + 3 a _{0}}) : y = - \frac{3}{( 5 + 3 a _{0} ) ^{2}} x + \frac{6 a _{0} + 5}{( 3 a _{0} + 5 ) ^{2}}

y = - \frac{3}{( 5 + 3 a _{0} ) ^{2}} x + \frac{6 a _{0} + 5}{( 3 a _{0} + 5 ) ^{2}}

What is the tangent of 1/(5+3x) ?
The tangent of 1/(5+3x) is y=-3/((5+3a_{0))^2}x+(6a_{0}+5)/((3a_{0)+5)^2}