Find #P'(t)# for #P(t) = 1/3(t^3 - 7t^2 + 6t +150)#?

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Answer 1 · 2018-02-20T11:51:52

#P'(t)=(3t^2-14t+6)/3#

We have to solve #d/dt1/3(t^3-7t^2+6t+150)#

As #d/dxaf#, where #a# is a constant, is equal to #ad/dxf#, we can write:

#1/3(d/dt(t^3-7t^2+6t+150))#

We can use the sum rule, which states that #d/dx(f+g)=d/dxf+d/dxg#. Now we can write:

#1/3(d/dtt^3-d/dt7t^2+d/dt6t+d/dt150)#

According to the power rule, #d/dxx^n=nx^(n-1)#. We can now solve:

#1/3(3t^(3-1)-7(2t^(2-1))+6t^(1-1)+0)#

#1/3(3t^2-14t+6)#

#(3t^2-14t+6)/3#, or:

#t^2+2-(14t)/3#