Results 1 to 5 of 5

Math Help - Derivative Help Please

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    3

    Derivative Help Please

    This is the problem I have f(x)= x^3 ( sec(x) )

    And this is what I think the answer is. Is it correct? If so could I possibly just distribute the sec(x) through (sec(x) tan(x) )? Thanks ahead of time.
    f'(x)= 6x^2 (sec(x)) (sec(x) tan(x))
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by CalcNewb View Post
    This is the problem I have f(x)= x^3 ( sec(x) )

    And this is what I think the answer is. Is it correct? If so could I possibly just distribute the sec(x) through (sec(x) tan(x) )? Thanks ahead of time.
    f'(x)= 6x^2 (sec(x)) (sec(x) tan(x))
    You need to apply product rule here.

    You should end up with f'(x)=3x^2\sec(x)+x^3\sec(x)\cdot\sec(x)\tan(x)=3x  ^2\sec(x)+x^3\sec^2(x)\tan(x)

    Does this make sense?

    --Chris
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    3
    Quote Originally Posted by Chris L T521 View Post
    You need to apply product rule here.

    You should end up with f'(x)=3x^2\sec(x)+x^3\sec(x)\cdot\sec(x)\tan(x)=3x  ^2\sec(x)+x^3\sec^2(x)\tan(x)

    Does this make sense?

    --Chris
    ok yes I do understand how to use the product rule. I used the product rule but got something totally different. Let me try to show you my steps.

    "first times derivative of second + second times derivative of the first" this is what our calc teacher told us to remember for product rule. so here it goes
    D=derivative

    x^3 D(sec(x)) + sec(x) D(x^3)
    so I get

    X^3 (sec(x) tan(x)) + sec(x) (3x^2)

    this is almost what you have I just dont see how you got the other sec^2(x)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Rhymes with Orange Chris L T521's Avatar
    Joined
    May 2008
    From
    Santa Cruz, CA
    Posts
    2,844
    Thanks
    3
    Quote Originally Posted by CalcNewb View Post
    ok yes I do understand how to use the product rule. I used the product rule but got something totally different. Let me try to show you my steps.

    "first times derivative of second + second times derivative of the first" this is what our calc teacher told us to remember for product rule. so here it goes
    D=derivative

    x^3 D(sec(x)) + sec(x) D(x^3)
    so I get

    X^3 (sec(x) tan(x)) + sec(x) (3x^2)

    this is almost what you have I just dont see how you got the other sec^2(x)
    Oh...woops!

    You're right...it should just be x^3\sec(x)\tan(x)+3x^2\sec(x)

    I was probably in a hurry and as a result, made a tiny mistake there.

    Sorry about that...

    --Chris
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    3
    Quote Originally Posted by Chris L T521 View Post
    Oh...woops!

    You're right...it should just be x^3\sec(x)\tan(x)+3x^2\sec(x)

    I was probably in a hurry and as a result, made a tiny mistake there.

    Sorry about that...

    --Chris
    whew I was getting worried..thank you
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. contuous weak derivative $\Rightarrow$ classic derivative ?
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: April 22nd 2011, 02:37 AM
  2. Replies: 0
    Last Post: January 24th 2011, 11:40 AM
  3. [SOLVED] Definition of Derivative/Alt. form of the derivative
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 23rd 2010, 06:33 AM
  4. Derivative Increasing ==> Derivative Continuous
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: February 23rd 2010, 10:58 AM
  5. Replies: 2
    Last Post: November 6th 2009, 02:51 PM

Search Tags


/mathhelpforum @mathhelpforum