derivatives

[dcgirl16]

im having a lot of trouble using the chain rule product rule and quotient rule..i can do them fine seperatly but when they're put together i cant get them like if you have (x^2-1)^4 (2-3x) i would start with
4(x^2-1)^3(2x)(2-3x)+(x^2-1)^4(-3)
have i done something wrong here because i never get the right answer with these ones i dont know if i messed up here or later on

[cristo]

You evaluation of the derivative is correct.

[dcgirl16]

ok so the next thing i would do would be to factor out the (x^2-1)to get
(8x)(x^2-1)^3(2-3x)+(x^2-1)(-3)
is this still right?

[dcgirl16]

i should have a 1 in there before the plus sign too

[cristo]

No. you've not factored out the (x^2-1), since you have (x^2-1)^3 in the first term. You're second term is incorrect since it should still be (x^2-1)^4. If you take out (x^2-1)^3 from both terms, and put it outside a bracket at the front, what do the remaining terms in the bracket look like?

[dcgirl16]

would it be
((x^2-1)^3)(1)(8X)(2-3X)+(x^2-1)(-3) ?

[cristo]

You're missing a bracket.. write it like this, omitting the unnecessary ones:
(x^2-1)^3(8x(2-3x)-3(x^2-1)).

Now, can you simplify this?

[dcgirl16]

my first thought would be to multiple the 8x and -3 through their brackets then combine like terms?

[cristo]

my first thought would be to multiple the 8x and -3 through their brackets then combine like terms?

Good idea. Give it a go and see what you come up with.

[dcgirl16]

ok that workedi got the right answer thanks for your help

[cristo]

You're very welcome!