# 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 :

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

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.

Note by Samuel Hatin
4 years, 10 months ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

• bulleted
• list

1. numbered
2. list

1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

Your 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.

- 4 years, 10 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!

- 4 years, 10 months ago