How do you use the chain rule to differentiate $y=(1/(t-3))^2$ ?

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Answer 1 · 2016-11-12T19:27:44

$dy/dt = -2/( t - 3 )^3$

Let $u = 1/ (t - 3)$

Therefore $y = u ^2$

Differentiate u with respect to t
$( du)/dt = - 1/(t-3)^2$

Differentiate y with respect to u
$dy/(du) = 2 u$

Using chain rule $dy/(dt) = dy/(du) xx (du)/(dt)$

therefore
$dy/(dt) = 2u xx -1/ (t - 3)^2$

$hArr dy/(dt) = 2 (1/(t - 3)) xx -1 /(t-3)^2$

$hArr dy/(dt) = -2 / (t-3)^ 3$

How do you use the chain rule to differentiate y=(1/(t-3))^2y=(1/(t-3))^2?