Question

can you help me, what is the first derivative of 12t (sin 3t^2) cos(3t^2) ????

Answer 1

thats going to get pretty ugly lol but I would use substitution to make it all easy!

Answer 2

sorry, 12t (sin 3t^2) (cos3t^2)

Answer 3

you would have to use the product rule 3 times so i'd start by doing 12t*sin(3t^2) first

Answer 4

This result should work. It can be proven by Mathematical Induction
\[(xyz)' = x'yz+ xy'z + xyz'\]

Answer 5

so the the derivative of 12t*sin(3t^2)= 12sin(3t^2)+72tcos(3t^2) now use the product rule again for the result i just found multiplied by cos(3t^2) hope this helps, like i said it's going to get ugly

Answer 6

yess @jrn1007 , you already helped me. many thanks :D

@Kira_Yamato i think your formula make it easy, thank you so much :)

Answer 7

Welcome!

Answer 8

In general, if you need to differentiate a product of n functions. Differentiate one of them and keep the rest, and do the same thing for the rest of the functions until all the n functions have been differentiated. Then all you need to do is to sum them all up.
\[\left( \prod_{r=1}^{n} f_r (x)\right)' = f_1'(x)f_2(x)...f_n(x) + f_1(x)f_2'(x)...f_n(x) +...+ f_1(x)f_2(x)...f_n'(x)\]
This can be proven by Mathematical Induction for n≥3

Answer 9

Actually n≥2

Answer 10

yess, I understand now @Kira_Yamato, thank you :D

Answer 11

My pleasure.

Answer 12

my answer is \[12(\sin 3t ^{2})(\cos 3t ^{2}) + 72t ^{2}(\cos ^{2} 3t^2) - 72t^2(\sin^2 3t^2)\] , can you correct my answer @Kira_Yamato @jrn1007 ????