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

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

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tangent of 1/(x+2)

tangent of

\frac{1}{x + 2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

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

Find the slope of

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

Find the line with slope m=

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

and passing through

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

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

\int_{0}^{\frac{π}{3}} sec (x) d x

d er i v a t i v e 2 x + 7

x \to 1 lim (x^{4} - 1)

\frac{d}{d x} (x^{\frac{4}{3}} - x^{\frac{1}{3}})

\int e^{6 θ} sin (7 θ) d θ

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