What is the tangent of y= 1/(x+2),\at x=7 ?

The tangent of y= 1/(x+2),\at x=7 is y=-1/81 x+16/81

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tangent of y= 1/(x+2),\at x=7

tangent of

y = \frac{1}{x + 2}, at x = 7

y = - \frac{1}{81} x + \frac{16}{81}

Solution steps

Find the tangent point:

(7, \frac{1}{9})

Find the slope of

y = \frac{1}{x + 2} : \frac{d y}{d x} = - \frac{1}{( x + 2 ) ^{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}{81}

Find the line with slope m=

- \frac{1}{81}

and passing through

(7, \frac{1}{9}) : y = - \frac{1}{81} x + \frac{16}{81}

y = - \frac{1}{81} x + \frac{16}{81}

x \to 8 lim (9 x + ∣ x - 8 ∣)

t an g e n t f (x) = x^{2} - \frac{1}{x}, (1, 0)

\frac{d}{d x} (\frac{1}{9 - x ^{2} - y ^{2} - z ^{2}})

((x - 1)^{2})^{^{'}}

\frac{\partial}{\partial z} (7 z e^{x yz})

What is the tangent of y= 1/(x+2),\at x=7 ?
The tangent of y= 1/(x+2),\at x=7 is y=-1/81 x+16/81