Results 1 to 4 of 4

Math Help - Product Rule Question

  1. #1
    Junior Member
    Joined
    Oct 2009
    Posts
    57

    Product Rule Question

    Hi,

    I'm asked to differentiate f(x) = 3(x^2 + 2)(4x^2 - 5x^4) - 3.

    Using the product rule, I know that f'(x) = u'v + uv'.

    What I'm confused about, is what to set u and v to in the above function. Would u=3(x^2 + 2)(4x^2 - 5x^4) and v = -3?

    Thanks,
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by centenial View Post
    Hi,

    I'm asked to differentiate f(x) = 3(x^2 + 2)(4x^2 - 5x^4) - 3.

    Using the product rule, I know that f'(x) = u'v + uv'.

    What I'm confused about, is what to set u and v to in the above function. Would u=3(x^2 + 2)(4x^2 - 5x^4) and v = -3?

    Thanks,
    No, not at all.

    u and v are functions of x.

    Since the derivative of a constant is zero, you can forget about the -3 at the end.

    Also the multiplier 3 can be used as simply an external factor.

    Hence

    u(x)=x^2+2

    v(x)=4x^2-5x^4

    \frac{d}{dx}\left[3\left(x^2+2\right)\left(4x^2-5x^4\right)-3\right]=3\frac{d}{dx}\left[\left(x^2+2\right)\left(4x^2-5x^4\right)\right]

    You could of course just multiply out the expression and differentiate term by term also.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    skeeter's Avatar
    Joined
    Jun 2008
    From
    North Texas
    Posts
    11,863
    Thanks
    639
    Quote Originally Posted by centenial View Post
    Hi,

    I'm asked to differentiate f(x) = 3(x^2 + 2)(4x^2 - 5x^4) - 3.

    Using the product rule, I know that f'(x) = u'v + uv'.

    What I'm confused about, is what to set u and v to in the above function. Would u=3(x^2 + 2)(4x^2 - 5x^4) and v = -3?

    Thanks,
    product rule ...

    u = 3(x^2+2)

    v = (4x^2-5x^4)

    the -3 term at the end of the expression is inconsequential since the derivative of -3 is 0
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Oct 2009
    Posts
    57
    Thanks! I think I understand now.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] quick question on product rule and equality rule for logs
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: October 19th 2011, 07:29 PM
  2. Product Rule Question
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: December 15th 2009, 01:32 AM
  3. Replies: 5
    Last Post: October 19th 2009, 01:04 PM
  4. Product Rule Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 7th 2009, 10:57 PM
  5. Sum/Product Rule Question
    Posted in the Discrete Math Forum
    Replies: 1
    Last Post: April 7th 2009, 07:16 PM

Search Tags


/mathhelpforum @mathhelpforum