Implicit Derivative using regular derivatives and partial derivatives

Today, I had a strange idea while doing implicit derivatives. Is there a way to find an implicit derivative using partial derivatives? I did some examples with functions and I studied the result with an exception but I'm struggling on the part where I want to prove that it woks for all kind of functions. Here's what I've come up with :

xy3+ysinx=x2+cosyx{y^3} + y\sin x = {x^2} + \cos y

With the regular implicit derivative I obtain :

dydx=2xy3ycosx3xy2+sinx+siny\frac{{dy}}{{dx}} = \frac{{2x - {y^3} - y\cos x}}{{3x{y^2} + \sin x + \sin y}}

With partial I decided to put z as a function of x and y then find the derivative with respect to both variables and then use an identity to get yx\frac{{\partial y}}{{\partial x}}

z=x2+cosyxy3ysinxzy=siny3xy2sinxzx=2xy3ycosxyx=yzzxyx=2xy3ycosxsiny3xy2sinx\begin{array}{l} z = {x^2} + \cos y - x{y^3} - y\sin x\\ \frac{{\partial z}}{{\partial y}} = - \sin y - 3x{y^2} - \sin x\\ \frac{{\partial z}}{{\partial x}} = 2x - {y^3} - y\cos x\\ \frac{{\partial y}}{{\partial x}} = \frac{{\partial y}}{{\partial z}}\frac{{\partial z}}{{\partial x}}\\ \frac{{\partial y}}{{\partial x}} = \frac{{2x - {y^3} - y\cos x}}{{ - \sin y - 3x{y^2} - \sin x}} \end{array}

Then I see that the only difference is a -1. I thought that maybe this is true:

yx=dydx - \frac{{\partial y}}{{\partial x}} = \frac{{dy}}{{dx}}

For the method I used.For an equation of this type I found that it holds:

f(x)=g(y)df(x)dx=dydxdg(y)dydydx=dydf(x)dxdg(y)z=g(y)f(x)zx=df(x)dxzy=dg(y)dyyx=dydf(x)dxdg(y)yx=dydx\begin{array}{l} f(x) = g(y)\\ \frac{{df(x)}}{{dx}} = \frac{{dy}}{{dx}}\frac{{dg(y)}}{{dy}}\\ \frac{{dy}}{{dx}} = \frac{{dydf(x)}}{{dxdg(y)}}\\ z = g(y) - f(x)\\ \frac{{\partial z}}{{\partial x}} = - \frac{{df(x)}}{{dx}}\\ \frac{{\partial z}}{{\partial y}} = \frac{{dg(y)}}{{dy}}\\ \frac{{\partial y}}{{\partial x}} = - \frac{{dydf(x)}}{{dxdg(y)}}\\ - \frac{{\partial y}}{{\partial x}} = \frac{{dy}}{{dx}} \end{array}

I've never followed any courses on partial so maybe some operations I did in this are not correct. I would really like to know if it's true, and if not then maybe where did I go wrong.

Note by Samuel Hatin
6 years, 8 months ago

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Your problem lies with the line where you write yx=yzzx \frac{\partial y}{\partial x} = \frac{\partial y}{\partial z}\, \frac{\partial z}{\partial x} Once you have functions of more than one variable, the chain rule becomes more complicated.

For example, if we have z=z(x,y)z = z(x,y) and we can write x=x(u1,,um)x = x(u_1,\ldots,u_m) and y=y(u1,,um)y = y(u_1,\ldots,u_m) as functions of u1,,umu_1,\ldots,u_m, then zz can be regarded as a function of u1,,umu_1,\ldots,u_m. The chain rule tells us that zuj=zxxuj+zyyuj \begin{array}{rcl} \frac{\partial z}{\partial u_j} & = & \frac{\partial z}{\partial x} \frac{\partial x}{\partial u_j} + \frac{\partial z}{\partial y} \frac{\partial y}{\partial u_j} \end{array} for any 1jm1 \le j \le m.

The m=1m=1 case of the chain rule works for you here. We are looking at an equation of the form z(x,y)  =  0 z(x,y) \; = \; 0 where yy is a function of xx. This means that zz can be regarded as a function of xx alone as well as a function of x,yx,y. Since x,yx,y are both functions of xx alone, partial derivatives can be replaced by ordinary derivatives (there are no other variables to keep constant!), and so xx  =  dxdx  =  1yx  =  dydx \frac{\partial x}{\partial x} \; = \; \frac{d x}{d x} \; = \; 1 \qquad \frac{\partial y}{\partial x} \; = \; \frac{d y}{d x} The chain rule now tells us that 0  =  dzdx  =  zxxx+zyyx  =  zx+zydydx() 0 \; = \; \frac{d z}{d x} \; = \; \frac{\partial z}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial x} \; = \; \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}\frac{d y}{d x} \qquad \qquad (\star) and hence we deduce that dydx  =  zx(zy)1 \frac{d y}{d x} \; = \; - \frac{\partial z}{\partial x}\left(\frac{\partial z}{\partial y}\right)^{-1} and the minus sign appears automatically. What is going on in equation ()(\star) is what you naturally write down when performing implicit differentiation.

Mark Hennings - 6 years, 8 months ago

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Thanks a lot! I didn't the chain rule had to be modified. This makes it way more clear and now I see why I'm obtain these results. Thanks again!

Samuel Hatin - 6 years, 8 months ago

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