Неявні функції

So far, almost all of the functions that we've seen so far are of the form $y=f(x)$. Examples include $y=3x$ or $y=-\sqrt{x+2}$. These are called explicit functions.

On the other hand, implicit functions, like their name suggests, are functions of the form $f(x_1,x_2,\ldots,x_n)=0$, where $x_1,x_2,\ldots, x_n$ are different variables. While this general expression may look scary, don't worry! In single-variable calculus, we only deal with implicit functions of the form $f(x,y)=0$.

So, what do these implicit functions look like? One example is $$xy-1=0.$$ Of course, for this equation we can solve for $y$ to get $$y=\frac{1}{x}.$$ By doing so, we changed our initially-implicit function to an explicit function. On the other hand, the implicit function $$x^2+y^2-1=0$$ can't be rearranged into one explicit function. The best we can do is solve for $y$ to get $$y=\pm\sqrt{1-x^2}.$$ The $\pm$ must be there because $$\left(\sqrt{1-x^2}\right)^2=\left(-\sqrt{1-x^2}\right)^2=1-x^2.$$ So, at best we can rewrite this implicit function as two separate explicit functions. Graphically, the two explicit functions are shown below:

Now that you've learned what implicit functions are, you can move on to implicit differentiation!

Here is some practice for implicit functions. Good luck!

(Videos) An Introduction to Implicit Differentiation

by PatrickJMT

Here's a video by patrickJMT explaining what an implicit equation is and how you go about solving it.

Before we begin let's look at the difference between an explicit equation and an implicit equation. An explicit equation is one where $y$ is by itself and $x$ is on the other side. $$y=x^2\\ y=\sin x+x$$ An implicit equation is where the $x$'s and $y$'s are all mixed up on the same side. $$x^2+xy=y\\ xy=x+y$$

Normally when taking a derivative we take it with respect to $x$. $$\frac{d}{dx}(x^2)=2x$$ Notice that the symbols match up. We have a $dx$ and an $x^2$.

But what if we take the derivative with respect to $x$ of a function that doesn't have an $x$? $$\frac{d}{dx}(y^2)$$ The symbols don't match up. But we would like them to. What we do is rewrite $\frac{d}{dx}$ as $\frac{d}{dy}\frac{dy}{dx}$. Now our expression becomes: $$\frac{d}{dy}\frac{dy}{dx}y^2$$ We can rearrange this: $$\frac{d}{dy}y^2\frac{dy}{dx}$$ Notice that the symbols match for $dy$ and $y^2$ now. So we can differentiate. $$2y\frac{dy}{dx}$$

This becomes the gist of how we solve these. We take the derivative like normal, when it's respect to $x$, and if we have something different, like a $y$ or $z$, we tack on a $\frac{dy}{dx}$ or $\frac{dz}{dx}$.

The procedure for solving these is simple.

1. Take the derivative and add $\frac{dy}{dx}$ where needed.
2. Get rid of parentheses.
3. Solve for $\frac{dy}{dx}$.

Let's look at an example now. Find $\frac{dy}{dx}$. $$x^2+xy+\cos y=8y$$ We start by taking the derivative of each piece. So, the derivative of $x^2$ is just $2x$. But what about $xy$? This is a product so we need to use the product rule. $$\color{red}{(x)'}(y)+(x)\color{red}{(y)'}\\ y+x\frac{dy}{dx}$$ For $\cos y$ we need to apply the chain rule. $$-\sin y\cdot\color{red}{(y)'}\\ -\sin y\frac{dy}{dx}$$ Finally, $8y$ becomes $8\frac{dy}{dx}$.

Putting all these together we get: $$2x+\left(y+x\frac{dy}{dx}\right)-\sin y\frac{dy}{dx}=8\frac{dy}{dx}$$

So step 2 tells us to get rid of the parentheses which is easy. $$2x+y+x\frac{dy}{dx}-\sin y\frac{dy}{dx}=8\frac{dy}{dx}$$

So now we have three terms with $\frac{dy}{dx}$ and two terms without. What we want to do is move all the terms with it to one side and all terms without it to the other side. Then we can factor out a $\frac{dy}{dx}$ and solve for it. $$2x+y+x\frac{dy}{dx}-\sin y\frac{dy}{dx}=8\frac{dy}{dx}\\ x\frac{dy}{dx}-\sin y\frac{dy}{dx}-8\frac{dy}{dx}=-2x-y\\ \frac{dy}{dx}(x-\sin y-8)=-2x-y\\ \boxed{\frac{dy}{dx}=\frac{-2x-y}{x-\sin y-8}}$$

Here's another video by Patrick where he shows you how to find the second derivative of an implicit equation. When finding higher power derivatives for implicit equations you usually end up plugging back in the expression from the previous derivatives.

So there's the first derivative, but now we need to find the second. To do this we need to use the quotient rule. $$\frac{d^2y}{dx^2}=\frac{\color{red}{(-9x)'}(y)-(-9x)\color{red}{(y)'}}{y^2}\\ \frac{d^2y}{dx^2}=\frac{-9y+9x\color{red}{\frac{dy}{dx}}}{y^2}$$ So what do we do with this $\frac{dy}{dx}$ in our problem now? Well, we actually have an expression for this since it's just the first derivative. $$\color{red}{\frac{dy}{dx}=-9\frac{x}y}\\ \frac{d^2y}{dx^2}=\frac{-9y+9x\color{red}{(\frac{-9x}{y})}}{y^2}\\ \frac{d^2y}{dx^2}=\frac{-9y-\frac{81x^2}{y}}{y^2}\\ \frac{d^2y}{dx^2}=\frac{-\frac{9y^2}{y}-\frac{81x^2}{y}}{y^2}\\ \frac{d^2y}{dx^2}=\frac{\frac{-9y^2-81x^2}{y}}{y^2}\\ \frac{d^2y}{dx^2}=\frac{-9y^2-81x^2}{y^3}\\ \frac{d^2y}{dx^2}=\frac{-9y^2-81x^2}{y^3}\\ \frac{d^2y}{dx^2}=\frac{-9(y^2+9x^2)}{y^3}\\$$ Now take a look at the expression in the parentheses, $y^2+9x^2$. This is the same expression that we started with above. $$\color{red}{x^2+y^2=9}$$ So we can substitute in $9$ for that! $$\frac{d^2y}{dx^2}=\frac{-9(\color{red}{9})}{y^3}\\ \boxed{\frac{d^2y}{dx^2}=-\frac{81}{y^3}}$$