Question

How do I find the derivative of ((t-2)^2)(t-4).. ?

Answer 1

product rule+chain rule

Answer 2

\[[(t-2)^2(t-4)]'=[(t-2)^2]'(t-4)+(t-2)^2[(t-4)]'\]

Answer 3

product rule!

Answer 4

\[2(t-2)^1(t-2)'(t-4)+(t-2)^2(1-0)\]

Answer 5

chain rule

Answer 6

\[2(t-2)(1-0)(t-4)+(t-2)^2(1)\]

Answer 7

\[2(t-2)(1)(t+4)+(t-2)^2\]

Answer 8

\[2(t-2)(t+4)+(t-2)^2\]

Answer 9

made a slight typo at the end, should be \[\large 2(t-2)(t-4)+(t-2)^2\] (should be t-4 not t+4) but other than that, it's perfect

Answer 10

optionally, you can expand and simplify to get \[\large 3t^2-16t+20\]

Answer 11

minus 4 plus 4 whats the difference lol
nice eye jim

Answer 12

the difference is t-4...literally lol