Question

differentiate the given function:
f(x)=\[(3x^3+4x)(x-5)(x+1)\]

Answer 1

12x^3-24*x^2-62x-20

Answer 2

that is not it.

Answer 3

15x^4-48x^3-33*x^2-32*x-20

Answer 4

is that correct?

Answer 5

yes it is. not really sure how you got that. can you explain it please?

Answer 6

were using this equation to find primes, is that correct?

Answer 7

sure
(3x^3+4*x)*(x-5)(x+1)
multiply
(3x^3+4*x)*(x-5) = 3x^4 - 15x^3 + 4x^2 - 20x
take the previous
(3x^4 - 15x^3 + 4x^2 - 20x) (x+1)
3x^5 - 15x^4 + 4x^3 - 20x^2 - 3x^4 - 15x^3 + 4x^2 - 20x
just add them then diff

Answer 8

@surdawi, that is an incorrect method,

USE PRODUCT RULE

Answer 9

and chain rule if applicable

Answer 10

hey psi, can i say that f(x)=(3x^3+4x) ?

Answer 11

and then find the prime of that?

Answer 12

so f prime would be 9x^2+4?

Answer 13

nope keep f(x) as it is , i.e. as given in the question

to find the derivative you have

f'(x)= (6x+4)*(x-5)(x+1)+(3x^2+4x) * (1-0)*(x+1)+(3x^2+4x) *(x-5)*(1+0)

Answer 14

we used chain and product rule

we find derivatives of each function at a time
i.e.DER OF (3x^2+4x)=6x+4, multiplied BY OTHER ORIGINAL FUNCTION
DER OF (X-5)=1, MULTIPLIED BY OTHER ORIGINAL FUNCTIONS
DER OF (X+1)=1, MULTIPLIED BY OTHER ORIGINAL FUNCITONS

Answer 15

@psi9epsilon, you still will get my answer 15x^4-48x^3-33*x^2-32*x-20

Answer 16

@surdawi, certainly we will still get the answer, but you need to learn how to solve the problem correctly and efficiently : ), this matters most when you are helping someone