How do you find the derivative of #f(t)=(4-t)^3#?

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Answer 1 · 2018-03-07T12:22:34

#f(t)=(-3)(4-t)^2#

#d/dt=3*(4-t)^2*d/dx(4-x)#
=#3*(4-t^2)*-1#
=#(-3)(4-t)^2#

Answer 2 · 2018-03-07T12:34:20

#-3(4-t)^2#

According to the chain rule, #(df)/dx=(df)/(du)*(du)/dx#, where #u# is a function within #f#.

Here, #f=u^3# where #u=4-t#, so we have:

#d/(du)u^3*d/dx(4-x)#

#3u^2*-1#

#-3u^2#, but since #u=4-t#, we have:

#-3(4-t)^2#