What is the tangent of f(x)=(sqrt(x))/(x+3),\at x=4 ?

The tangent of f(x)=(sqrt(x))/(x+3),\at x=4 is y=-1/196 x+15/49

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tangent of f(x)=(sqrt(x))/(x+3),\at x=4

tangent of

f (x) = \frac{x}{x + 3}, at x = 4

y = - \frac{1}{196} x + \frac{15}{49}

Solution steps

Find the tangent point:

(4, \frac{2}{7})

Find the slope of

f (x) = \frac{x}{x + 3} : \frac{df ( x )}{d x} = \frac{- x + 3}{2 x ( x + 3 ) ^{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{1}{196}

Find the line with slope m=

- \frac{1}{196}

and passing through

(4, \frac{2}{7}) : y = - \frac{1}{196} x + \frac{15}{49}

y = - \frac{1}{196} x + \frac{15}{49}

\frac{1}{s ( s - 2 ) ^{2}}

y = \frac{x}{64 + x ^{2}}, (0, 0)

\frac{d}{d x} (2 sin (x) - 1)

\int (ln (x y)) d x

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

What is the tangent of f(x)=(sqrt(x))/(x+3),\at x=4 ?
The tangent of f(x)=(sqrt(x))/(x+3),\at x=4 is y=-1/196 x+15/49