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

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

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tangent of y=(2x-1)/(x+2)

tangent of

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

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

Find the slope of

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

Compute the slope of

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

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

Find the line with slope m=

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

and passing through

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

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

\frac{d}{d x} (x^{2} (ln (x)))

\frac{d}{d x} (ln (6 sec (x)))

f (x) = 2 x^{2} - 3 x + 1, 1, 2

y = 9 x, y = x^{2}

\int \frac{25 x ^{2} - 9}{x ^{3}} d x

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