What is the tangent of 3/x+5/(x^3),\at x=-1 ?

The tangent of 3/x+5/(x^3),\at x=-1 is y=-18x-26

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tangent of 3/x+5/(x^3),\at x=-1

tangent of

\frac{3}{x} + \frac{5}{x ^{3}}, at x = - 1

y = - 18 x - 26

Solution steps

Find the tangent point:

(- 1, - 8)

Find the slope of

y = \frac{3}{x} + \frac{5}{x ^{3}} : \frac{d y}{d x} = - \frac{3}{x ^{2}} - \frac{15}{x ^{4}}

Compute the slope of

y = \frac{3}{x} + \frac{5}{x ^{3}}

x = - 1 : m = - 18

Find the line with slope m=

- 18

and passing through

(- 1, - 8) : y = - 18 x - 26

y = - 18 x - 26

\frac{d}{d x} (\frac{x}{x - 3})

\frac{\partial}{\partial x} (x^{2} + y^{2} + z^{2})

\int \frac{2}{( 1 - x ) ^{2}} d x

ln (1 + 2 x), 0

\frac{cos ^{2} ( 2 θ )}{cos ( θ )}

What is the tangent of 3/x+5/(x^3),\at x=-1 ?
The tangent of 3/x+5/(x^3),\at x=-1 is y=-18x-26