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

The tangent of y= 1/(5+2x),(-1, 1/3) is y=-2/9 x+1/9

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

tangent of

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

y = - \frac{2}{9} x + \frac{1}{9}

Solution steps

Find the slope of

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

Compute the slope of

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

(- 1, \frac{1}{3}) : m = - \frac{2}{9}

Find the line with slope m=

- \frac{2}{9}

and passing through

(- 1, \frac{1}{3}) : y = - \frac{2}{9} x + \frac{1}{9}

y = - \frac{2}{9} x + \frac{1}{9}

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

\int - t d t

x \to 3 lim (x (\frac{2}{x} + 5))

\int_{\frac{1}{7}}^{5} 12 x ln (7 x) d x

y = e^{- 4 x}

What is the tangent of y= 1/(5+2x),(-1, 1/3) ?
The tangent of y= 1/(5+2x),(-1, 1/3) is y=-2/9 x+1/9