Results 1 to 5 of 5

Math Help - f'(x)

  1. #1
    Newbie
    Joined
    Oct 2009
    Posts
    2

    f'(x)

    f'(x) X/(x^2-4)^(1/2)





    hi!! im stuck on this and dont understand how to find its derivitive. ive tried multipling by the conjugate and i cant get anything to cancel. it is an ap test question that my teacher gave us for homework.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Amer's Avatar
    Joined
    May 2009
    From
    Jordan
    Posts
    1,093
    Quote Originally Posted by bromez View Post
    f'(x) X/(x^2-4)^(1/2)





    hi!! im stuck on this and dont understand how to find its derivitive. ive tried multipling by the conjugate and i cant get anything to cancel. it is an ap test question that my teacher gave us for homework.
    I will not solve it I will solve a similar example

    f(x) = \frac{\sin x }{(x^3-1)^{\frac{3}{2}}}

    I prefer to write it like this

    f(x) = (\sin x)(x^3-1)^{\frac{-3}{2}}

    f'(x)= (x^3-1)^{\frac{-3}{2}}\left(\frac{d \sin x}{dx}\right) + (\sin x)\left(\frac{d(x^3-1)^{\frac{-3}{2}}}{dx}\right)

    f'(x) = (x^3-1)^{\frac{3}{2}}(\cos x) + (\sin x)\left(\frac{-3}{2}(3x^2)(x^3-1)^{\frac{-5}{2}}\right)

    in general

    f(x) = g(x) h(x)

    f'(x) = g'(x) h(x) + h'(x) g(x)
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Quote Originally Posted by bromez View Post
    f'(x) X/(x^2-4)^(1/2)





    hi!! im stuck on this and dont understand how to find its derivitive. ive tried multipling by the conjugate and i cant get anything to cancel. it is an ap test question that my teacher gave us for homework.
    f'(x)=x(x^2 - 4)^{-\frac{1}{2}}
    Use product and chain rule y'=vu' + uv' where v=(x^2 - 4)^{-\frac{1}{2}} and u=x
    u'=1 and v'=-\frac{1}{2}(x^2 - 4)^{-\frac{3}{2}}(2x)
    Substitute these values into the formula above.
    f'(x)=(x^2 -4)^{-\frac{1}{2}}(1) + (x)(-x(x^2 -4)^{-\frac{3}{2}})
    Last edited by VitaX; October 16th 2009 at 12:50 AM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2009
    Posts
    2
    thx this should get me started
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member VitaX's Avatar
    Joined
    Feb 2009
    From
    Ohio
    Posts
    185
    Had an error with v' above. Fixed it though
    Follow Math Help Forum on Facebook and Google+

Search Tags


/mathhelpforum @mathhelpforum