Question

derivative of y= (x^4+3x^2-2)^5

Answer 1

you us ethe chain rule - are you familiar with this?

Answer 2

f'(u^n) = nu'u(n-1)

F'(x) = 5 (4x^3 + 6x) (x^4 + 3x^2 -2)^4

Answer 3

[f(g(x)]' = f'g(x) * g'(x)

Answer 4

but the answer in the bookis diff

Answer 5

10 x (3+2 x^2) (-2+3 x^2+x^4)^4

Answer 6

i got your asnwers, but i dont undestnad how that becomes what the book says

Answer 7

5(4x^3+6x)^4

(20x^3+30x)^4
(5*(x^4+3x^2-2)^4)*(4x^3+6x)

Answer 8

jordan's answer is correct but it can be simplidied

10x(2x^2 + 3)(x^4 + 3x^2 - 2)^4

Answer 9

oh okay

Answer 10

sorry about that

Answer 11

@cwrw238 how do u simplify it. can you do it step by step. i dont see how it becomes that final answer

Answer 12

the first part can be simplified

5(4x^3 + 6x)

take the part in the parentheses
the greatest common factor is 2x
right?

Answer 13

oh. ya, okay.

Answer 14

so we can write it as 2x(2x^2 + 3)

Answer 15

this multiplied by the 5 gives
10x(2x^2 + 3)

Answer 16

thank you so much for your help!

Answer 17

yw