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 :

\[x{y^3} + y\sin x = {x^2} + \cos y\]

With the regular implicit derivative I obtain :

\[\frac{{dy}}{{dx}} = \frac{{2x - {y^3} - y\cos x}}{{3x{y^2} + \sin x + \sin y}}\]

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 \[\frac{{\partial y}}{{\partial x}}\]

\[\begin{array}{l} z = {x^2} + \cos y - x{y^3} - y\sin x\\ \frac{{\partial z}}{{\partial y}} = - \sin y - 3x{y^2} - \sin x\\ \frac{{\partial z}}{{\partial x}} = 2x - {y^3} - y\cos x\\ \frac{{\partial y}}{{\partial x}} = \frac{{\partial y}}{{\partial z}}\frac{{\partial z}}{{\partial x}}\\ \frac{{\partial y}}{{\partial x}} = \frac{{2x - {y^3} - y\cos x}}{{ - \sin y - 3x{y^2} - \sin x}} \end{array}\]

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

\[ - \frac{{\partial y}}{{\partial x}} = \frac{{dy}}{{dx}}\]

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

\[\begin{array}{l} f(x) = g(y)\\ \frac{{df(x)}}{{dx}} = \frac{{dy}}{{dx}}\frac{{dg(y)}}{{dy}}\\ \frac{{dy}}{{dx}} = \frac{{dydf(x)}}{{dxdg(y)}}\\ z = g(y) - f(x)\\ \frac{{\partial z}}{{\partial x}} = - \frac{{df(x)}}{{dx}}\\ \frac{{\partial z}}{{\partial y}} = \frac{{dg(y)}}{{dy}}\\ \frac{{\partial y}}{{\partial x}} = - \frac{{dydf(x)}}{{dxdg(y)}}\\ - \frac{{\partial y}}{{\partial x}} = \frac{{dy}}{{dx}} \end{array}\]

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.

## Comments

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

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

The \(m=1\) case of the chain rule works for you here. We are looking at an equation of the form \[ z(x,y) \; = \; 0 \] where \(y\) is a function of \(x\). This means that \(z\) can be regarded as a function of \(x\) alone as well as a function of \(x,y\). Since \(x,y\) are both functions of \(x\) alone, partial derivatives can be replaced by ordinary derivatives (there are no other variables to keep constant!), and so \[ \frac{\partial x}{\partial x} \; = \; \frac{d x}{d x} \; = \; 1 \qquad \frac{\partial y}{\partial x} \; = \; \frac{d y}{d x} \] The chain rule now tells us that \[ 0 \; = \; \frac{d z}{d x} \; = \; \frac{\partial z}{\partial x}\frac{\partial x}{\partial x} + \frac{\partial z}{\partial y}\frac{\partial y}{\partial x} \; = \; \frac{\partial z}{\partial x} + \frac{\partial z}{\partial y}\frac{d y}{d x} \qquad \qquad (\star) \] and hence we deduce that \[ \frac{d y}{d x} \; = \; - \frac{\partial z}{\partial x}\left(\frac{\partial z}{\partial y}\right)^{-1} \] 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 · 3 years, 9 months ago

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– Samuel Hatin · 3 years, 9 months ago

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!Log in to reply