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#