Results 1 to 6 of 6

Thread: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

  1. #1
    Newbie
    Joined
    Aug 2011
    Posts
    2

    Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

    Hello,

    I just graduated and now feel masochistically inclined to torture myself with diff eq. My question relates to differentiating terms in implicit functions using the product rule. I understand how to use the product rule with two differentiable functions of x. I do not, however, quite understand how to use this rule when the terms involve y and x.

    I have attached a screenshot of the problem. I have highlighted in red the term whose origin I do not quite understand.

    f(x) = y^4 + 2xy + 6x = 7

    u = 2xy
    v = y

    The derivative of 2xy w.r.t x is 2y(dy/dx). But why?

    Question: Why does dy/dx appear from u(dv/dx)? Why is the derivative not simply 2y?

    Thank you for any help,
    postg
    Attached Thumbnails Attached Thumbnails Finding Derivatives in Implicit Functions with Two Variables (Product Rule)-untitled.png  
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6

    Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

    In my opinion what is not full explained is the fact that y isn't an independent variable, but is a function of x, so that if $\displaystyle u(x)=2\ x^{2}$ and $\displaystyle v(x)= y^{2}(x)$ is...

    $\displaystyle u\ \frac{d v}{d x} = 2\ x^{2}\ 2\ y(x)\ \frac{dy}{dx}= 4\ x^{2}\ y\ y^{'}$ (1)

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2011
    Posts
    2

    Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

    Why do you define v(x) = y(x)?
    The problem defines v = y, not v = y(x). What am I missing?

    I'm afraid I don't understand your answer to the question. Why is the derivative of y = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member TheChaz's Avatar
    Joined
    Nov 2010
    From
    Northwest Arkansas
    Posts
    600
    Thanks
    2

    Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

    Quote Originally Posted by postg View Post
    Why do you define v(x) = y(x)?
    The problem defines v = y, not v = y(x). What am I missing?

    I'm afraid I don't understand your answer to the question. Why is the derivative of y = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?
    There is also the CHAIN RULE.
    The derivative of (x^2), with respect to x, is 2x(dx/dx). dx/dx is ONE.
    The derivative of (y^2), with respect to x, is 2y(dy/dx). dy/dx is NOT ONE!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    12,880
    Thanks
    1946

    Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

    Quote Originally Posted by postg View Post
    Why do you define v(x) = y(x)?
    The problem defines v = y, not v = y(x). What am I missing?

    I'm afraid I don't understand your answer to the question. Why is the derivative of y = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?
    It's because $\displaystyle \displaystyle y$ is a function of $\displaystyle \displaystyle x$. The derivative is taken with respect to $\displaystyle \displaystyle x$, not $\displaystyle \displaystyle y$.

    You want to find the derivative of $\displaystyle \displaystyle v = [y(x)]^2$, in other words $\displaystyle \displaystyle \frac{dv}{dx}$.

    Let $\displaystyle \displaystyle w = y(x)$ so that $\displaystyle \displaystyle v = w^2$. From the chain rule, we would know that $\displaystyle \displaystyle \frac{dv}{dx} = \frac{dw}{dx} \cdot \frac{dv}{dw}$.

    $\displaystyle \displaystyle \frac{dw}{dx} = \frac{d}{dx}[y(x)] = \frac{dy}{dx}$.

    $\displaystyle \displaystyle \frac{dv}{dw} = 2w = 2y$.

    So $\displaystyle \displaystyle \frac{dv}{dx} = 2y\frac{dy}{dx}$.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,782
    Thanks
    2824
    Awards
    1

    Re: Finding Derivatives in Implicit Functions with Two Variables (Product Rule)

    Quote Originally Posted by postg View Post
    Why is the derivative of y = 2y(dy/dx). Why not just 2y, by the nx^n-1 rule?
    Look at the word implicit. When asking for the derivative of $\displaystyle y^2$ with respect to $\displaystyle x$, we are implying (it is implicit) that $\displaystyle y^2$ is a function $\displaystyle x$,

    Thus $\displaystyle \frac{d(xy^2)}{dx}=y^2+2xy\frac{dy}{dx}.$
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. product rule, taking derivatives
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Jan 5th 2010, 11:06 AM
  2. Derivatives - product rule + chain rule
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Oct 19th 2009, 03:26 PM
  3. Replies: 1
    Last Post: Oct 5th 2009, 07:22 AM
  4. Derivatives - product rule with three terms?
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Mar 25th 2009, 12:06 PM
  5. Replies: 3
    Last Post: Mar 20th 2009, 03:46 AM

Search tags for this page

Click on a term to search for related topics.

Search Tags


/mathhelpforum @mathhelpforum