What is the derivative of 8x+8x^{-1} ?

The derivative of 8x+8x^{-1} is 8-8/(x^2)

What is the first derivative of 8x+8x^{-1} ?

The first derivative of 8x+8x^{-1} is 8-8/(x^2)

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derivative of

8 x + 8 x^{- 1}

8 - \frac{8}{x ^{2}}

Solution steps

\frac{d}{d x} (8 x + 8 x^{- 1})

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{d}{d x} (8 x) + \frac{d}{d x} (8 x^{- 1})

\frac{d}{d x} (8 x) = 8

\frac{d}{d x} (8 x^{- 1}) = - \frac{8}{x ^{2}}

= 8 - \frac{8}{x ^{2}}

\frac{d}{d x} (e^{- \frac{x}{y}})

\frac{d}{d x} (\frac{4}{3} p i x^{3})

3 x^{2}, at x = 4

x \to 5 lim (x)

t \to + 0 lim (\frac{8}{t} - \frac{8}{t ^{2} + t})

What is the derivative of 8x+8x^{-1} ?
The derivative of 8x+8x^{-1} is 8-8/(x^2)
What is the first derivative of 8x+8x^{-1} ?
The first derivative of 8x+8x^{-1} is 8-8/(x^2)