What is the tangent of 6x^{-3},\at x=5 ?

The tangent of 6x^{-3},\at x=5 is y=-18/625 x+24/125

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tangent of 6x^{-3},\at x=5

tangent of

6 x^{- 3}, at x = 5

y = - \frac{18}{625} x + \frac{24}{125}

Solution steps

Find the tangent point:

(5, \frac{6}{125})

Find the slope of

y = 6 x^{- 3} : \frac{d y}{d x} = - \frac{18}{x ^{4}}

Compute the slope of

y = 6 x^{- 3}

x = 5 : m = - \frac{18}{625}

Find the line with slope m=

- \frac{18}{625}

and passing through

(5, \frac{6}{125}) : y = - \frac{18}{625} x + \frac{24}{125}

y = - \frac{18}{625} x + \frac{24}{125}

\frac{x - y}{x + y}

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

\int_{0}^{1} 2 x^{2} - 1 d x

\frac{d x}{d t} = 2 x (5 - x)

y^{^{'}} = e^{2 y + 6 x}

What is the tangent of 6x^{-3},\at x=5 ?
The tangent of 6x^{-3},\at x=5 is y=-18/625 x+24/125