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

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    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
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    MHF Contributor chisigma's Avatar
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    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 u(x)=2\ x^{2} and v(x)= y^{2}(x) is...

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

    Kind regards

    \chi \sigma
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    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?
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    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!
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    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 y is a function of \displaystyle x. The derivative is taken with respect to \displaystyle x, not \displaystyle y.

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

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

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

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

    So \displaystyle \frac{dv}{dx} = 2y\frac{dy}{dx}.
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    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 y^2 with respect to x, we are implying (it is implicit) that y^2 is a function x,

    Thus \frac{d(xy^2)}{dx}=y^2+2xy\frac{dy}{dx}.
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