How do you differentiate $x + \frac{4}{x}$ $x+4/x$ ?

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How do you differentiate $x + \frac{4}{x}$ $x+4/x$ ?

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Answer 1 · 2016-06-08T10:43:06

$1 - \frac{4}{x^{2}}$ $1-4/x^2$

Explanation:

Begin by expressing $x + \frac{4}{x} = x + 4 x^{- 1}$ $x+4/x=x+4x^-1$

now differentiate using the $power rule$ $color(blue)"power rule"$

$\Rightarrow \frac{d}{d x} (x + 4 x^{- 1}) = 1 - 4 x^{- 2} = 1 - \frac{4}{x^{2}}$ $rArrd/dx(x+4x^-1)=1-4x^-2=1-4/x^2$

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How do you differentiate $x + \frac{4}{x}$ $x+4/x$ ?

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