Results 1 to 5 of 5

Math Help - Implicit differentiation word problem

  1. #1
    Newbie
    Joined
    Sep 2011
    Posts
    2

    Question Implicit differentiation word problem

    I have a written assignment in which one specific word problem is confusing me greatly. I just recently learned implicit differentiation but haven't gotten much into depth with it. The question is as follows:

    "Each time we use implicit differentiation, we are able to rewrite the resulting equation in the form:

    f(x,y)=y′ ·g(x,y)

    for some expressions f and g.

    Explain why this can always be done; that is, why doesn’t the chain rule ever produce a term like:

    (y′)^2

    or

    1/y′ "

    I really am not grasping the form of the initial equation given let alone the logic behind what it is asking. I don't want anyone to give me the answer to the problem, but any help in understanding what it is actually asking, the concepts behind the question and the thought process required to answer the question would be of great help.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    44

    Re: Implicit differentiation word problem

    F(x,y(x))=0\Rightarrow \frac {\partial F}{\partial x}+\frac {\partial F}{\partial y}y'(x)=0\Rightarrow \ldots
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    10,969
    Thanks
    1009

    Re: Implicit differentiation word problem

    Quote Originally Posted by elrpsu View Post
    I have a written assignment in which one specific word problem is confusing me greatly. I just recently learned implicit differentiation but haven't gotten much into depth with it. The question is as follows:

    "Each time we use implicit differentiation, we are able to rewrite the resulting equation in the form:

    f(x,y)=y′ ·g(x,y)

    for some expressions f and g.

    Explain why this can always be done; that is, why doesn’t the chain rule ever produce a term like:

    (y′)^2

    or

    1/y′ "

    I really am not grasping the form of the initial equation given let alone the logic behind what it is asking. I don't want anyone to give me the answer to the problem, but any help in understanding what it is actually asking, the concepts behind the question and the thought process required to answer the question would be of great help.
    Every function of y is a function of x. So if you wanted to differentiate something like \displaystyle z= y^2 (in other words, \displaystyle z = \left[y(x)\right]^2 )with respect to x, you need to use the chain rule.

    The "inner" function is \displaystyle y(x), and the "outer" function is \displaystyle y^2.

    So using the chain rule, if \displaystyle z = y^2 and we let \displaystyle u = y \implies z = u^2, then

    \displaystyle \frac{du}{dx} = \frac{dy}{dx} and \displaystyle \frac{dz}{du} = 2u = 2y.

    Therefore \displaystyle \frac{dz}{dx}= \frac{dz}{du} \cdot \frac{du}{dx} = 2y\,\frac{dy}{dx}.


    So whenever you want to differentiate a function of y, since y is the inner function, you are always going to get \displaystyle \frac{dy}{dx} as a factor because of the chain rule.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Sep 2011
    Posts
    2

    Re: Implicit differentiation word problem

    Quote Originally Posted by Prove It View Post
    Every function of y is a function of x. So if you wanted to differentiate something like \displaystyle z= y^2 (in other words, \displaystyle z = \left[y(x)\right]^2 )with respect to x, you need to use the chain rule.

    The "inner" function is \displaystyle y(x), and the "outer" function is \displaystyle y^2.

    So using the chain rule, if \displaystyle z = y^2 and we let \displaystyle u = y \implies z = u^2, then

    \displaystyle \frac{du}{dx} = \frac{dy}{dx} and \displaystyle \frac{dz}{du} = 2u = 2y.

    Therefore \displaystyle \frac{dz}{dx}= \frac{dz}{du} \cdot \frac{du}{dx} = 2y\,\frac{dy}{dx}.


    So whenever you want to differentiate a function of y, since y is the inner function, you are always going to get \displaystyle \frac{dy}{dx} as a factor because of the chain rule.
    Thank you! That helped me understand this all quite a bit better than I was able to before.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    14,973
    Thanks
    1121

    Re: Implicit differentiation word problem

    Another way of expressing it is that differentiation is a linear operation: (f+ g)'= f'+ g' and (fg)'= f'g+ fg'. You never get "squares" or more complicated functions of the derivative.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Implicit Differentiation word problem
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 4th 2010, 10:13 PM
  2. word problem using implicit differentiation
    Posted in the Calculus Forum
    Replies: 2
    Last Post: February 16th 2010, 10:41 AM
  3. implicit differentiation word problem
    Posted in the Calculus Forum
    Replies: 3
    Last Post: April 4th 2009, 03:58 PM
  4. word problem using implicit differentiation
    Posted in the Calculus Forum
    Replies: 1
    Last Post: November 28th 2008, 01:01 PM
  5. Implicit differentiation word problem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 19th 2008, 01:54 PM

Search Tags


/mathhelpforum @mathhelpforum