What is the tangent of 2/(1-t),\at t=-2 ?

The tangent of 2/(1-t),\at t=-2 is y= 2/9 t+10/9

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tangent of 2/(1-t),\at t=-2

tangent of

\frac{2}{1 - t}, at t = - 2

y = \frac{2}{9} t + \frac{10}{9}

Solution steps

Find the tangent point:

(- 2, \frac{2}{3})

Find the slope of

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

Compute the slope of

y = \frac{2}{1 - t}

t = - 2 : m = \frac{2}{9}

Find the line with slope m=

\frac{2}{9}

and passing through

(- 2, \frac{2}{3}) : y = \frac{2}{9} t + \frac{10}{9}

y = \frac{2}{9} t + \frac{10}{9}

\frac{d}{d x} (2 x ln (2 x) + x)

\frac{d y}{d x} - \frac{2 x}{2 - x ^{2}} \cdot y = e^{x}, y (2) = 4

f (x) = 3 0^{3} - 3 x^{2} + 12 x - 74

\frac{d}{d x} (6 e^{- 2 x})

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

What is the tangent of 2/(1-t),\at t=-2 ?
The tangent of 2/(1-t),\at t=-2 is y= 2/9 t+10/9