How do you differentiate $y = \frac{1}{{(t - 1)}^{2}}$ $y=1/(t-1)^2$ ?

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How do you differentiate $y = \frac{1}{{(t - 1)}^{2}}$ $y=1/(t-1)^2$ ?

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Answer 1 · 2017-03-05T11:43:23

$\frac{d y}{d t} = - \frac{2}{{(t - 1)}^{3}}$ $dy/dt = -2/(t-1)^3$

Explanation:

We have:

$y = \frac{1}{{(t - 1)}^{2}} = {(t - 1)}^{- 2}$ $y = 1/(t-1)^2 = (t-1)^(-2)$

So then by the chain rule, we have:

$\frac{d y}{d t} = (- 2) {(t - 1)}^{- 3} \cdot (1)$ $dy/dt = (-2)(t-1)^(-3)*(1)$
$= - 2 {(t - 1)}^{- 3}$ $" " = -2(t-1)^(-3)$
$= - \frac{2}{{(t - 1)}^{3}}$ $" " = -2/(t-1)^3$

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How do you differentiate $y = \frac{1}{{(t - 1)}^{2}}$ $y=1/(t-1)^2$ ?

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How do you differentiate y=1/(t-1)^2y=1(t−1)2y=1/(t-1)^2?

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Explanation:

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How do you differentiate $y = \frac{1}{{(t - 1)}^{2}}$ $y=1/(t-1)^2$ ?