What is the tangent of y=(x^2)/(x+3),(2, 4/5) ?

The tangent of y=(x^2)/(x+3),(2, 4/5) is y= 16/25 x-12/25

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tangent of y=(x^2)/(x+3),(2, 4/5)

tangent of

y = \frac{x ^{2}}{x + 3}, (2, \frac{4}{5})

y = \frac{16}{25} x - \frac{12}{25}

Solution steps

Find the slope of

y = \frac{x ^{2}}{x + 3} : \frac{d y}{d x} = \frac{x ^{2} + 6 x}{( x + 3 ) ^{2}}

Compute the slope of

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

(2, \frac{4}{5}) : m = \frac{16}{25}

Find the line with slope m=

\frac{16}{25}

and passing through

(2, \frac{4}{5}) : y = \frac{16}{25} x - \frac{12}{25}

y = \frac{16}{25} x - \frac{12}{25}

f (x) = 4 x^{2} + 1

y^{^{'}} + \frac{1}{2} y = 8, y (0) = 2

\frac{\partial}{\partial x} (\frac{( x v ^{2} )}{( x ^{2} + v ^{2} )})

(\frac{x + 2}{x}) (7 - 2 x^{3})

\int \frac{5}{( 5 x + 4 ) ln ( 5 x + 4 )} d x

What is the tangent of y=(x^2)/(x+3),(2, 4/5) ?
The tangent of y=(x^2)/(x+3),(2, 4/5) is y= 16/25 x-12/25