What is the tangent of y= 1/(x^5),(3, 1/243) ?

The tangent of y= 1/(x^5),(3, 1/243) is y=-5/729 x+2/81

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tangent of y= 1/(x^5),(3, 1/243)

tangent of

y = \frac{1}{x ^{5}}, (3, \frac{1}{243})

y = - \frac{5}{729} x + \frac{2}{81}

Solution steps

Find the slope of

y = \frac{1}{x ^{5}} : \frac{d y}{d x} = - \frac{5}{x ^{6}}

Compute the slope of

y = \frac{1}{x ^{5}}

(3, \frac{1}{243}) : m = - \frac{5}{729}

Find the line with slope m=

- \frac{5}{729}

and passing through

(3, \frac{1}{243}) : y = - \frac{5}{729} x + \frac{2}{81}

y = - \frac{5}{729} x + \frac{2}{81}

\int_{- 3}^{3} 18 - 2 x^{2} d x

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

(x y + y^{2} + x^{2}) d x - x^{2} d y = 0

x \to 0 + lim (\frac{1}{x})

x \to - 4 lim (- (\frac{6}{x + 4}))

What is the tangent of y= 1/(x^5),(3, 1/243) ?
The tangent of y= 1/(x^5),(3, 1/243) is y=-5/729 x+2/81