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\frac{d}{dx}\left(\frac{1}{\left(x-1\right)^2}\right)

Derive the function 1/((x-1)^2) with respect to x

Answer

$\frac{-\frac{d}{dx}\left(1-2x+x^2\right)}{\left(x-1\right)^{4}}$

Step-by-step explanation

Problem

$\frac{d}{dx}\left(\frac{1}{\left(x-1\right)^2}\right)$
1

Applying the quotient rule which states that if $f(x)$ and $g(x)$ are functions and $h(x)$ is the function defined by ${\displaystyle h(x) = \frac{f(x)}{g(x)}}$, where ${g(x) \neq 0}$, then ${\displaystyle h'(x) = \frac{f'(x) \cdot g(x) - g'(x) \cdot f(x)}{g(x)^2}}$

$\frac{\left(x-1\right)^2\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(\left(x-1\right)^2\right)}{\left(\left(x-1\right)^2\right)^2}$

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Answer

$\frac{-\frac{d}{dx}\left(1-2x+x^2\right)}{\left(x-1\right)^{4}}$

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$\frac{d}{dx}\left(\frac{1}{\left(x-1\right)^2}\right)$

Main topic:

Differential calculus

Used formulas:

2. See formulas

Time to solve it:

0.33 seconds