Question

what is the first derivative of -5t^3 - t^2(5t+2)^3 + (8t^2 - 2t) ????

Answer 1

-15t^2 -2t(5t+2)^3 + 15*(5t+2)^2 + 16t -2

Answer 2

remember that \[\dfrac{d}{dx}(u\cdot v)=u\cdot \dfrac{dv}{dx}+v\cdot \dfrac{du}{dx}.\]

Answer 3

\[\sin ^{2} (3t ^{2}) = 2(\sin 3t ^{2})( \cos 3t ^{2})(6t)\] in first derivative?

Answer 4

yes

Answer 5

and how about \[-5t ^{2} \cos (2t ^{3})\] ???

Answer 6

-5t^2(-sin (2t^2))(4t) the negatives cancel out

Answer 7

wait is that a 3 or a 2?

Answer 8

-10t(cos 2t^3)(-sin 2t^3)(6t^2), how about this?

Answer 9

u have to multiply that out, there's no way u can use the product rule here...

Answer 10

what is the difference between −5t^2cos(2t^3) and sin^2(3t^2)???

Answer 11

well..one's cosine and one's sine lol what are u asking?

Answer 12

I mean, why there's no way I can use the product rule in −5t^2cos(2t^3)

Answer 13

yes u can for this one

Answer 14

derivative of −5t^2 times cos(2t^3) + the derivative of cos(2t^3) times −5t^2

Answer 15

hmm, i see thankz @11calcBC

Answer 16

np