What is the tangent of y=-1/(2x^3),(2,-1/16) ?

The tangent of y=-1/(2x^3),(2,-1/16) is y= 3/32 x-1/4

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tangent of y=-1/(2x^3),(2,-1/16)

tangent of

y = - \frac{1}{2 x ^{3}}, (2, - \frac{1}{16})

y = \frac{3}{32} x - \frac{1}{4}

Solution steps

Find the slope of

y = - \frac{1}{2 x ^{3}} : \frac{d y}{d x} = \frac{3}{2 x ^{4}}

Compute the slope of

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

(2, - \frac{1}{16}) : m = \frac{3}{32}

Find the line with slope m=

\frac{3}{32}

and passing through

(2, - \frac{1}{16}) : y = \frac{3}{32} x - \frac{1}{4}

y = \frac{3}{32} x - \frac{1}{4}

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

\frac{d}{d x} (\frac{3 x + 6}{2 x + 3})

y^{^{''}} + 80 y^{^{'}} + 1200 y = 0

\frac{\partial}{\partial x} (4 cos (3 x))

\int 2 x (4 x - 3)^{3} d x

What is the tangent of y=-1/(2x^3),(2,-1/16) ?
The tangent of y=-1/(2x^3),(2,-1/16) is y= 3/32 x-1/4