What is the tangent of y= 1/(2+3x),(2, 1/8) ?

The tangent of y= 1/(2+3x),(2, 1/8) is y=-3/64 x+7/32

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tangent of y= 1/(2+3x),(2, 1/8)

tangent of

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

y = - \frac{3}{64} x + \frac{7}{32}

Solution steps

Find the slope of

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

Compute the slope of

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

(2, \frac{1}{8}) : m = - \frac{3}{64}

Find the line with slope m=

- \frac{3}{64}

and passing through

(2, \frac{1}{8}) : y = - \frac{3}{64} x + \frac{7}{32}

y = - \frac{3}{64} x + \frac{7}{32}

\frac{d}{d x} (\frac{4}{x ^{2} + 1})

x \to 0 lim (\frac{1}{2 x} - \frac{1}{e ^{2 x} - 1})

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

7 x - 2 x

y = 20 x^{3} (x^{4} + 5)^{4}

What is the tangent of y= 1/(2+3x),(2, 1/8) ?
The tangent of y= 1/(2+3x),(2, 1/8) is y=-3/64 x+7/32