What is the tangent of y=(sqrt(x))/(x+1),(4, 2/5) ?

The tangent of y=(sqrt(x))/(x+1),(4, 2/5) is y=-3/100 x+13/25

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tangent of y=(sqrt(x))/(x+1),(4, 2/5)

tangent of

y = \frac{x}{x + 1}, (4, \frac{2}{5})

y = - \frac{3}{100} x + \frac{13}{25}

Solution steps

Find the slope of

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

Compute the slope of

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

(4, \frac{2}{5}) : m = - \frac{3}{100}

Find the line with slope m=

- \frac{3}{100}

and passing through

(4, \frac{2}{5}) : y = - \frac{3}{100} x + \frac{13}{25}

y = - \frac{3}{100} x + \frac{13}{25}

x \to 0 lim (cos (x^{2}))

7 x - x^{2}, 2 x

f (x) = \frac{1}{2} (x + 1)^{- \frac{1}{2}}

n = 1 \sum \infty (\frac{1}{5})^{n}

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

What is the tangent of y=(sqrt(x))/(x+1),(4, 2/5) ?
The tangent of y=(sqrt(x))/(x+1),(4, 2/5) is y=-3/100 x+13/25