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:

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

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