# Thread: derivative of function using accurate notation

1. ## derivative of function using accurate notation

hey everyone, so the question is y= -5 over x cubed minus 2 over x + (x+1)

for -5 over x cubed:using quotient rule, let u= 5 v= x cubed u'= 0 v'= 1

y'(x)= x cubed(0) - 5(1) all over (x+1)^2

= 0-5 all over (x+1)^2 ?????

am I on the right track? or am i supposed to include the minus sign somwhere?

2. ## Re: derivative of function using accurate notation

Hi,

I do not undestand why you put: v = x cubed and v´= 1

derivative of x cubed = derivative of (x^3) = 3x^2

Let me know....

Thanks

3. ## Re: derivative of function using accurate notation

or yea, thats right. my bad

4. ## Re: derivative of function using accurate notation

my main problem is with the 1st fraction and the negative sign in front of it. what do i do?

5. ## Re: derivative of function using accurate notation

Ok,

Using (u/v)´= (u´v-uv´)/v^2

and taking u = -5 and v = x^3

derivative of -5/(x^3) is: [0(x^3)-(-5)3x^2]/(x^3)^2 =

15x^2/x^6

= 15/x^4

Let me know if you need more help

6. ## Re: derivative of function using accurate notation

ok so from the part with [0(x^3)-(-5)3x^2]/(x^3)^2... can u explain how to expand (-5)3x^2 as in the steps please?

7. ## Re: derivative of function using accurate notation

oh wait my bad, i meant from the 15x^2/(x^3)^2

8. ## Re: derivative of function using accurate notation

ok (x^3)^2 = (x^3)(x^3) = x^6

Then: 15x^2/(x^3)^2 = 15x^2/x^6

If you simplify (divding by x^2)

The result is 15/x^4

9. ## Re: derivative of function using accurate notation

ohhhh....ook thanks thanks thanks (y) was having problems there

10. ## Re: derivative of function using accurate notation

so finally, in solving the original question would it be y'= (15/x^4)- (2/x^2) + 2x + 2?