Results 1 to 5 of 5

Math Help - Differentiation using the Composite Rule

  1. #1
    Newbie
    Joined
    Sep 2008
    Posts
    3

    Differentiation using the Composite Rule

    Hi, I'm new to this forum and am abit stuck.

    I need to differentiate the function f(x)=10+x(squared) - this function is also a square root but I dont know how to input that here.

    What I have so far is y=sq.root u, where u = 10+x(squared)
    then dy/du = 1/2 10/sq.rootu and du/dx=2x
    I think the composite rule says dy/dx = dy/du du/dx = 1/2 1/sq.root u. 2x = x/sq.root 10+x(squared). Hopefully this makes sense!?

    I then have to use the quotient rule together with my answer above to show that the function g(x)= (e to the power of x/7)/sq.root 10 + x(squared) has the derivative

    g'(x) = e to the power of x/7(xsquard - 7x + 10) divided by 7(10 + x squared) to the power of 3/2
    I will be extremely grateful for any help.
    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2008
    Posts
    1,092
    Your first differentiation is correct. The derivative of \sqrt{10 + x^2} is indeed \frac{x}{\sqrt{10 + x^2}}.

    For the second problem, the derivative of \frac{f(x)}{g(x)} is \frac{f(x)g'(x) - g(x)f'(x)}{(g(x))^2}.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Quote Originally Posted by honeybumble View Post
    Hi, I'm new to this forum and am abit stuck.

    I need to differentiate the function f(x)=10+x(squared) - this function is also a square root but I dont know how to input that here.
    f^2(x)=10+x^2
    If you don't want to use latex, use ^ for exponents, and parentheses please

    What I have so far is y=\sqrt{u}, where u=10+x^2
    then \frac{dy}{du} =\frac 12 \times \frac{1}{\sqrt{u}} and \frac{du}{dx}=2x
    Correct so far
    I think the composite rule says \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}
    Yes it does !
    \frac 12 \times \frac{1}{\sqrt{u}} \times 2x=\frac{x}{\sqrt{10+x^2}}. Hopefully this makes sense!?
    Yes.

    Now, a problem with that. If f^2(x)=10+x^2 then f(x)=\sqrt{10+x^2} OR f(x)=-\sqrt{10+x^2} !!!
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2008
    Posts
    3
    Thanks for your help! It is much appreciated.

    I have further problems I would be grateful of further clarification.

    Firstly, how do I find any stationary points of the function g(x) as defined by the quotient rule component of my previous post. I also need to use the First Derivative Test to classify each stationary point as a local maximum or local minimum of g(x).

    Also what are the rules to find the general solution of the differential equation (using the answer to the f(x) function in my first post) of the differential equation

    dy/du = 10+y(squared) / y (x>0, y>0) - sorry I cant work out where to find the "squared" and "square root" symbols on this page....

    giving the answer in implicit form

    and then find the solution of the differential equation for which y + sq.root6 when x=0 giving this particular solution in explicit form.

    Many thanks
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member 11rdc11's Avatar
    Joined
    Jul 2007
    From
    New Orleans
    Posts
    894
    To find your stationary points equal your derivative to 0 and see what your x values are. To use the first derivative test you then have to see in what interval the sign changes from positive to negative or vice versa. Here is an example, let me know if you need me to explain some more.

    http://www.mathhelpforum.com/math-he...tive-test.html
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Composite Rule
    Posted in the Calculus Forum
    Replies: 8
    Last Post: September 19th 2009, 05:26 AM
  2. Composite rule help
    Posted in the Calculus Forum
    Replies: 2
    Last Post: August 17th 2009, 10:33 AM
  3. using the composite rule???
    Posted in the Calculus Forum
    Replies: 1
    Last Post: May 31st 2009, 05:04 AM
  4. Please help with Composite Rule
    Posted in the Calculus Forum
    Replies: 4
    Last Post: June 7th 2008, 01:12 PM
  5. Composite Rule problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: May 23rd 2008, 01:43 AM

Search Tags


/mathhelpforum @mathhelpforum