What is the tangent of (x^2)/(x+8) ?

The tangent of (x^2)/(x+8) is y=(a_{0}^2+16a_{0})/((a_{0)+8)^2}x-(8a_{0}^2)/((a_{0)+8)^2}

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tangent of (x^2)/(x+8)

tangent of

\frac{x ^{2}}{x + 8}

y = \frac{a _{0}^{2} + 16 a _{0}}{( a _{0} + 8 ) ^{2}} x - \frac{8 a _{0}^{2}}{( a _{0} + 8 ) ^{2}}

Solution steps

Compute the tangent line to the general point

x = a_{0}

Find the tangent point:

(a_{0}, \frac{a _{0}^{2}}{a _{0} + 8})

Find the slope of

y = \frac{x ^{2}}{x + 8} : \frac{d y}{d x} = \frac{x ^{2} + 16 x}{( x + 8 ) ^{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{a _{0}^{2} + 16 a _{0}}{( a _{0} + 8 ) ^{2}}

Find the line with slope m=

\frac{a _{0}^{2} + 16 a _{0}}{( a _{0} + 8 ) ^{2}}

and passing through

(a_{0}, \frac{a _{0}^{2}}{a _{0} + 8}) : y = \frac{a _{0}^{2} + 16 a _{0}}{( a _{0} + 8 ) ^{2}} x - \frac{8 a _{0}^{2}}{( a _{0} + 8 ) ^{2}}

y = \frac{a _{0}^{2} + 16 a _{0}}{( a _{0} + 8 ) ^{2}} x - \frac{8 a _{0}^{2}}{( a _{0} + 8 ) ^{2}}

x^{^{'}} = e^{x}

\int 12 x^{2} d x

\int - 8 tan^{2} (x) d x

\frac{d}{d x} (- 6 sin (\frac{π}{12} x) + 18)

\int_{0}^{\frac{π}{2}} cos^{5} (y) d y

What is the tangent of (x^2)/(x+8) ?
The tangent of (x^2)/(x+8) is y=(a_{0}^2+16a_{0})/((a_{0)+8)^2}x-(8a_{0}^2)/((a_{0)+8)^2}