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Posted

I recently heard of a different kind of implicit differentiation, apparently it works like this:

 

[math]\frac{dy}{dx} = - \frac{dt}{dx} \div \frac{dt}{dy}[/math]

 

Where t is a made-up variable. This makes complete sense to me except for the negative sign. Why is there a negative sign and what is the name of this method?

  • 3 weeks later...
Posted

Let us assume that the equation [math]t(x,y)=0[/math] defines [math]y[/math] as an implicit function of [math]x[/math]. Why implicit? Because we cannot always easily express y as a function of x from the equation for t(x,y). For example,[math]c^3+y^3+3axy=0[/math], where c,a are constants. In order, to get y as a function of x we must solve cubic equation (difficult).

Now without trying to find the dependence of y on x explicitly, let's just call the solution [math]y=y(x)[/math].

Now, constant 0 (of t(x,y)=0) may be thought of as a function of x, depending on it both directly and through [math]y(x)[/math]. Taking derivative we have:

[math]\frac{d t}{d x}=\frac{\partial t}{\partial x}+\frac{\partial t}{\partial y} \frac{d y}{d x} [/math]

Rearranging terms we have

[math] \frac{d y}{d x}=-\frac{\frac{\partial t}{\partial x}}{\frac{\partial t}{\partial y}}[/math]

Thus, we have found the derivative [math]\frac{d y}{d x}[/math] without finding y(x).

So negative sign appears because of the constant in the RHS of the definition of the implicit function t(x,y)=const.

Hope this answered your question.

Also see Wikepedia: http://en.wikipedia.org/wiki/Implicit_function

Posted

Yea me 2, in like 4 posts ive read of yours, i already know that your going to be on my imaginary ranking of scienceforums.net intellects, up with matt grime, atheist, swansont, woelen and ajb. trust me, if anyone had all their brains combined, heisenberg was a dwarf.

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