What is the slope of x^{-5}(1+x^{-1}),\at x=1 ?

The slope of x^{-5}(1+x^{-1}),\at x=1 is -11

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slope of x^{-5}(1+x^{-1}),\at x=1

slope

x^{- 5} (1 + x^{- 1}), at x = 1

- 11

Solution steps

y = x^{- 5} (1 + x^{- 1})

In order to find the slope of the function, take the derivative of

x^{- 5} (1 + x^{- 1})

\frac{d}{d x} (x^{- 5} (1 + x^{- 1})) = \frac{- 5 x - 6}{x ^{7}}

\frac{- 5 x - 6}{x ^{7}}

Find the value of

\frac{- 5 x - 6}{x ^{7}}

for

x = 1 : - 11

- 11

y = x^{2}, y^{2} = x

f (x) = 5 x - x^{2}

y^{^{''}} - 10 y + 2 y = 0

x^{^{'}} + 3 x = 0

\frac{\partial}{\partial x} (x^{6} y - 6 x^{2} y^{2})

What is the slope of x^{-5}(1+x^{-1}),\at x=1 ?
The slope of x^{-5}(1+x^{-1}),\at x=1 is -11