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I Implicit differentiation

  1. Sep 29, 2016 #1
    First of all thanks for the help, i have a problem finding a good explanation of de ecuation (d/dx)f=(∂f/∂x)+(∂f/∂y)*(dy/dx) could anyone write me a good explanation of this ecuation? thanks for the help
     
  2. jcsd
  3. Sep 29, 2016 #2

    andrewkirk

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    This is only true if ##f## is a function of two variables and there is a function ##g## of one variable such that ##y=g(x)##.

    Then we have ##f(x,y)=f(x,g(x))## so then, using the definitions of derivatives as limits, we can write:
    \begin{align*}
    \frac d{dx}f(x,y)&=\frac d{dx}f(x,g(x))\\
    &=\lim_{h\to 0}\left[\frac{f(x+h,g(x+h))-f(x,g(x))}{h}\right]\\
    &=\lim_{h\to 0}\left[
    \frac{f(x+h,g(x+h))-f(x,g(x+h))}{h}
    +\frac{f(x,g(x+h))-f(x,g(x))}{h}
    \right]\\
    &=\lim_{h\to 0}\left[
    \frac{f(x+h,g(x+h))-f(x,g(x+h))}{h}
    \right]+
    \lim_{h\to 0}\left[
    \frac{f(x,g(x+h))-f(x,g(x))}{h}
    \right]
    \end{align*}
    provided both limits exist.

    The first term is a bit tricky. Provided certain assumptions hold, the details of which we won't go into here, it is equal to the double limit:
    \begin{align*}
    \lim_{h'\to 0}\bigg[
    \lim_{h\to 0}\bigg(
    &\frac{f(x+h,g(x+h'))-f(x,g(x+h'))}{h}
    \bigg)
    \bigg]\\
    &=\lim_{h'\to 0}\bigg[
    \frac{\partial}{\partial x}f(x,g(x+h'))
    \bigg]\\
    &=
    \frac{\partial}{\partial x}f(x,g(x))\\
    \end{align*}

    The second term is equal to:
    \begin{align*}
    \lim_{h\to 0}\bigg[
    \frac{f(x,g(x+h))-f(x,g(x))}{g(x+h)-g(x)}
    &\times
    \frac{g(x+h)g(x)}{h}
    \bigg]\\
    &=\lim_{h\to 0}\left[
    \frac{f(x,g(x+h))-f(x,g(x))}{g(x+h)-g(x)}\right]
    \times
    \lim_{h\to 0}\left[\frac{g(x+h)-g(x)}{h}
    \right]\\
    &=\lim_{\delta y\to 0}\left[
    \frac{f(x,y+\delta y)-f(x,y)}{\delta y}\right]
    \times
    \lim_{h\to 0}\left[\frac{g(x+h)-g(x)}{h}
    \right]\\
    &=
    \frac{\partial }{\partial y}f(x,y)
    \times
    \frac{dg}{dx}(x)\\
    &=
    \frac{\partial }{\partial y}f(x,y)
    \times
    \frac{dy}{dx}\\
    \end{align*}
    Summing the terms, we get
    $$\frac{d}{dx}f(x,y)=\frac{\partial}{\partial x}f(x,y)+\frac{\partial }{\partial y}f(x,y)
    \times
    \frac{dy}{dx}$$
    as required.
     
  4. Sep 30, 2016 #3
    thanks a lot for the help
     
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