What is the tangent of y=(6x)/(x+2) ?

The tangent of y=(6x)/(x+2) is y=(12)/((a_{0)+2)^2}x+(6a_{0}^2)/((a_{0)+2)^2}

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tangent of y=(6x)/(x+2)

tangent of

y = \frac{6 x}{x + 2}

y = \frac{12}{( a _{0} + 2 ) ^{2}} x + \frac{6 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{6 a _{0}}{a _{0} + 2})

Find the slope of

y = \frac{6 x}{x + 2} : \frac{d y}{d x} = \frac{12}{( x + 2 ) ^{2}}

Compute the slope of

y = \frac{6 x}{x + 2}

x = a_{0} : m = \frac{12}{( a _{0} + 2 ) ^{2}}

Find the line with slope m=

\frac{12}{( a _{0} + 2 ) ^{2}}

and passing through

(a_{0}, \frac{6 a _{0}}{a _{0} + 2}) : y = \frac{12}{( a _{0} + 2 ) ^{2}} x + \frac{6 a _{0}^{2}}{( a _{0} + 2 ) ^{2}}

y = \frac{12}{( a _{0} + 2 ) ^{2}} x + \frac{6 a _{0}^{2}}{( a _{0} + 2 ) ^{2}}

\frac{d}{d x} ((1 + 5 x) (1 - 7 x))

\frac{d}{d x} ((lo g_{e} (x))^{x})

y^{^{''}} + 7 y^{^{'}} + 10 y = 0, y (0) = 1, y^{^{'}} (0) = 0

x \to 0 lim (\frac{2}{3 + 4 ^{\frac{1}{x}}})

What is the tangent of y=(6x)/(x+2) ?
The tangent of y=(6x)/(x+2) is y=(12)/((a_{0)+2)^2}x+(6a_{0}^2)/((a_{0)+2)^2}