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

The tangent of y= 5/(x^2+1),(2,1) is y=-4/5 x+13/5

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

tangent of

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

y = - \frac{4}{5} x + \frac{13}{5}

Solution steps

Find the slope of

y = \frac{5}{x ^{2} + 1} : \frac{d y}{d x} = - \frac{10 x}{( x ^{2} + 1 ) ^{2}}

Compute the slope of

y = \frac{5}{x ^{2} + 1}

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

Find the line with slope m=

- \frac{4}{5}

and passing through

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

y = - \frac{4}{5} x + \frac{13}{5}

\int 7 x cos (8 x) d x

\int x cos (4 x) d x

x^{2} + 3 x + 9

n = 1 \sum \infty \frac{x ^{n}}{n + 3}

\frac{1}{2 s ^{3}}

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