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

The tangent of y= 2/(1-3x),\at x=-1 is y= 3/8 x+7/8

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tangent of y= 2/(1-3x),\at x=-1

tangent of

y = \frac{2}{1 - 3 x}, at x = - 1

y = \frac{3}{8} x + \frac{7}{8}

Solution steps

Find the tangent point:

(- 1, \frac{1}{2})

Find the slope of

y = \frac{2}{1 - 3 x} : \frac{d y}{d x} = \frac{6}{( 1 - 3 x ) ^{2}}

Compute the slope of

y = \frac{2}{1 - 3 x}

x = - 1 : m = \frac{3}{8}

Find the line with slope m=

\frac{3}{8}

and passing through

(- 1, \frac{1}{2}) : y = \frac{3}{8} x + \frac{7}{8}

y = \frac{3}{8} x + \frac{7}{8}

\int \frac{x + 8}{x ^{2} + 9} d x

\int \frac{1}{x ^{2} - 400} d x

\frac{\partial}{\partial x} (2 x^{8 y})

\int t^{4} ln (t) d t

f (x) = - \frac{11}{s ^{5}}

What is the tangent of y= 2/(1-3x),\at x=-1 ?
The tangent of y= 2/(1-3x),\at x=-1 is y= 3/8 x+7/8