How do you differentiate $f (x) = \frac{1}{x^{2}}$ $f(x)=1/x^2$ ?

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How do you differentiate $f (x) = \frac{1}{x^{2}}$ $f(x)=1/x^2$ ?

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Answer 1 · 2016-11-03T05:15:15

$\frac{d}{d x} f (x) = - \frac{2}{x^{3}}$ $d/dxf(x) = -2/ x^3$

Explanation:

$f (x) = \frac{1}{x^{2}}$ $f(x)= 1/x^2$
$\frac{d}{d x} f (x) = \frac{d}{d x} (\frac{1}{x^{2}})$ $d/dxf(x) =d/dx(1/x^2)$
$\frac{d}{d x} (\frac{1}{x^{2}}) = \frac{d}{d x} (x^{- 2})$ $d/dx(1/x^2) = d/dx(x^-2)$
$\Rightarrow \frac{d}{d x} f (x) = - 2 \cdot x^{- 3}$ $implies d/dxf(x) = -2* x^-3$
$\Rightarrow \frac{d}{d x} f (x) = - \frac{2}{x^{3}}$ $implies d/dxf(x) = -2/ x^3$

Answer 2 · 2016-11-03T05:22:54

$\frac{d y}{d x} = \frac{- 2}{x^{3}}$ $dy/dx=(-2)/x^3$

Explanation:

Given -

$y = \frac{1}{x^{2}}$ $y=1/x^2$

Method -1
The given function can be written as -

$y = x^{- 2}$ $y=x^-2$

$\frac{d y}{d x} = - 2 x^{- 2 - 1}$ $dy/dx=-2x^(-2-1)$

$\frac{d y}{d x} = - 2 x^{- 3}$ $dy/dx=-2x^-3$

$\frac{d y}{d x} = \frac{- 2}{x^{3}}$ $dy/dx=(-2)/x^3$

Method - 2
We can apply quotient rule

$\frac{d y}{d x} = \frac{[(x^{2}) (0)] - [(1) (2 x)]}{{(x^{2})}^{2}}$ $dy/dx={[(x^2)(0)]-[(1)(2x)]}/((x^2)^2)$

$\frac{d y}{d x} = \frac{- 2 x}{x^{4}}$ $dy/dx=(-2x)/x^4$

$\frac{d y}{d x} = \frac{- 2}{x^{3}}$ $dy/dx=(-2)/x^3$

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