What is the tangent of 3/(x+2) ?

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

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tangent of 3/(x+2)

tangent of

\frac{3}{x + 2}

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

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{3}{a _{0} + 2})

Find the slope of

y = \frac{3}{x + 2} : \frac{d y}{d x} = - \frac{3}{( x + 2 ) ^{2}}

Compute the slope of

y = \frac{3}{x + 2}

x = a_{0} : m = - \frac{3}{( a _{0} + 2 ) ^{2}}

Find the line with slope m=

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

and passing through

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

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

\int_{1}^{3} (4 x^{3} - 3 x^{2}) d x

\frac{1}{s ^{2} + s}

\frac{d ^{2} y}{d t ^{2}} - e^{2} y = 0

\int \frac{u}{u ^{2} + 2} d u

\int_{0}^{1} x + 4 d x

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