What is the tangent of y= 3/(x+1),(2,1) ?

The tangent of y= 3/(x+1),(2,1) is y=-1/3 x+5/3

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tangent of y= 3/(x+1),(2,1)

tangent of

y = \frac{3}{x + 1}, (2, 1)

y = - \frac{1}{3} x + \frac{5}{3}

Solution steps

Find the slope of

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

Compute the slope of

y = \frac{3}{x + 1}

(2, 1) : m = - \frac{1}{3}

Find the line with slope m=

- \frac{1}{3}

and passing through

(2, 1) : y = - \frac{1}{3} x + \frac{5}{3}

y = - \frac{1}{3} x + \frac{5}{3}

y^{^{''}} + y = sec^{2} (x)

x \to 2 lim (x^{3} + 1)

\frac{d y}{d x} = \frac{x ^{2} - y ^{2} + y}{x}

\frac{\partial}{\partial x} (\frac{1}{x ^{2} + 3 y})

\int \frac{x ^{2} + x}{x ^{3} - x ^{2} + x - 1} d x

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