What is the tangent of f(x)=((x^2))/(x+9),\at x=3 ?

The tangent of f(x)=((x^2))/(x+9),\at x=3 is y= 7/16 x-9/16

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tangent of f(x)=((x^2))/(x+9),\at x=3

tangent of

f (x) = \frac{( x ^{2} )}{x + 9}, at x = 3

y = \frac{7}{16} x - \frac{9}{16}

Solution steps

Find the tangent point:

(3, \frac{3}{4})

Find the slope of

f (x) = \frac{x ^{2}}{x + 9} : \frac{df ( x )}{d x} = \frac{x ^{2} + 18 x}{( x + 9 ) ^{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{7}{16}

Find the line with slope m=

\frac{7}{16}

and passing through

(3, \frac{3}{4}) : y = \frac{7}{16} x - \frac{9}{16}

y = \frac{7}{16} x - \frac{9}{16}

\int sin^{3} (x) cos^{7} (x) d x

y^{^{''}} + 4 π^{2} y = 0

\frac{d}{d x} (2000 (1 + \frac{x}{12})^{60})

\frac{d}{d x} ((cos (5 x))^{x})

\int \frac{1}{16 x ^{2} + 1} d x

What is the tangent of f(x)=((x^2))/(x+9),\at x=3 ?
The tangent of f(x)=((x^2))/(x+9),\at x=3 is y= 7/16 x-9/16