隐微分法

Implicit Differentiation

The Chain Rule States: $$\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$

Or

$$f(x)= g(h(x)) \implies f'(x) = g'(x) h'(g(x))$$

Example: $$\frac{d}{dx}\left(\cos (y)\right)$$

Hence, $\frac{d}{dx}\cos(y)$ can be rewritten as:

$$\frac{d}{dx}\left(\cos (y(x))\right)$$

Now, lets say: $g(x) = y(x)$ and $h(x)=cos(x)$

$$g'(x)=y'(x)=\frac{dy}{dx}$$ $$h'(g(x)) = -\sin{(y(x))}$$

Now, Through the Chain rule we have: $$\frac{d}{dx}\left(\cos (y(x))\right) = -\sin{(y(x))}\frac{dy}{dx}$$

Lets try a harder example!

Find $\frac{dy}{dx}$:

$$y^5 + 2y^4 -7y^3 + x^2 - x = 0 \\ y(x)^5 + 2y(x)^4 -7y(x)^3 + x^2 - x = 0 \\ \frac{d}{dx}\left(y(x)^5 + 2y(x)^4 -7y(x)^3 + x^2 - x \right)= 0 \\ 5y(x)^4\frac{dy}{dx} + 8y(x)^3\frac{dy}{dx} -21y(x)^2\frac{dy}{dx} + 2x - 1= 0 \\ \frac{dy}{dx}\left(5y(x)^4 + 8y(x)^3-21y(x)^2\right)= 1-2x \\ \frac{dy}{dx}= \frac{1-2x}{5y^4 + 8y^3-21y^2}$$

There! We did it!

Implicit Functions

In algebra or calculus, we call an equation implicit if it is of the form $\mathcal{R}\left(x_1,\cdots,x_n\right)=0$, where $\mathcal{R}$ is a function of several variables, often a polynomial. For example, the implicit equation of the unit circle is $x^2+y^2-1=0$.

The above is the most formal definition of an implicit equation. However, we can often rearrange terms in variable forms and call them all implicit forms. For example, an implicit equation of the unit circle could also be $x^2+y^2=1$.

Disclaimer: "implicit" does not necessarily mean the function does not have a "nice" explicit form. For instance, $f(x) = x^\pi$ is in "explicit" form, but it is also implicitly defined by the solution to $\log y - \pi \log(x) = 0$ for $y$ in terms of $x$. Similar for $x^2+y^2=1$ vs. $\sqrt{1-x^2}$.

Calculus of implicit functions

Let us say we had an implicit function $f(x,y)=c$, where $c$ is a constant, and we wanted to find the derivative (or rate of change) $\frac{\mathrm{d}y}{\mathrm{d}x}$.

Alright, let's do this with an example. Consider the implicit function $$x^2 + xy = 3.$$Let us differentiate each side of the equation with respect to $x$, not forgetting the chain rule. We get $$2x + x \cdot \frac {\mathrm{d}y}{\mathrm{d}x} + 1 \cdot y = 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = - \frac {y + 2x}{x}.$$

Great; we have successfully found the correct derivative! Sure, it is in terms of both the $x$-coordinate and the $y$-coordinate, but that doesn't make things harder. We just have to remember that $y = \frac {3-x^2}{x}$, and, in this case, we can even substitute back in to get: $$\frac{\mathrm{d}y}{\mathrm{d}x} = - \frac {\frac {3-x^2}{x} + 2x}{x} = - \frac {3}{x^2} - 1.$$This was a relatively easy example. We could do it for more variables and we would get the same type of result. We'll let you slog out one more.

Problem. Find $\frac{\mathrm{d}z}{\mathrm{d}y}$ if $$xy^2z^2 - xz + 2x = 3z + y.$$

We could, but we don't have to. Let's not.

We want a derivative with respect to $y$, so let's go! Don't forget all the derivative rules ... what do we get? Click Continue to verify your answer after you think you have it.

$$2xy^2 z \cdot \frac {\mathrm{d}z}{\mathrm{d}y} + 2xyz^2 + y^2z^2 - x \cdot \frac {\mathrm{d}z}{\mathrm{d}y} - \frac {\mathrm{d}x}{\mathrm{d}y} \cdot z + 2\cdot \frac{\mathrm{d}x}{\mathrm{d}y} = 3z + y.$$

This is just a matter of using the derivatives correctly. Note that both $\frac {\mathrm{d}x}{\mathrm{d}y}$ and $\frac {\mathrm{d}z}{\mathrm{d}y}$ are involved $-$ this should be the case! The rest is algebra.

See the Practice! section of this section to practice more implicit differentiation.

Here's an example:

Find the slope of the line tangent to the circle $x^2+y^2=25$ at point $(3,4)$.

We know we can do it geometrically... But not now. We want to differentiate, but it's not written in the $y=\cdots$ form.

You can do it by solving for $y$ in terms of $x$. You will get 2 different solutions, but you still can figure out which one point $(3,4)$ is on. But we want to present another way to do it (which could make things much easier in some much more complicated equations).

We know that $y$ is related to $x$ somehow, let's assume that there's a function that expressed $y$ in terms of $x$.

Now take the derivative with respect to $x$ to both sides.

So you have all three components: $2x+2yy'=0$. Solving it, we get $y'=-\frac{x}{y}$.

The form might seem surprising since it also involves $y$, but remember that in a circle, we cannot solely determine the value of $y$ when $x=3$! We need both values!

Now plug in both values.

You get it. This is your first experience with implicit differentiation! Enjoy!

(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.

$$9x^2+y^2=9\\ 18x+2y\frac{dy}{dx}=0\\ 2y\frac{dy}{dx}=-18x\\ \frac{dy}{dx}=-\frac{18x}{2y}\\ \frac{dy}{dx}=-9\frac{x}y$$

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! $$\frac{d^2y}{dx^2}=\frac{-9(\color{red}{9})}{y^3}\\ \boxed{\frac{d^2y}{dx^2}=-\frac{81}{y^3}}$$

(Videos) Implicit Differentiation Examples

by PatrickJMT

Here's a video by patrickJMT going over some basic examples of implicit differentiation and working through the examples.

The first one is finding $\frac{dy}{dx}$ for: $$x^{-2}-\ln y^3=7\\ -2x^{-3}-\frac{1}{y^3}\cdot\color{red}{(y^3)'}=0\\ -2x^{-3}-\frac{1}{y^3}\cdot3y^2\cdot\color{red}{(y)'}=0\\ -2x^{-3}-\frac1{y^3}(3y^2)\frac{dy}{dx}=0\\ -2x^{-3}-\frac3y\frac{dy}{dx}=0\\ -\frac3y\frac{dy}{dx}=2x^{-3}\\ \frac{dy}{dx}=(-\frac y3)\frac2{x^3}\\ \boxed{\frac{dy}{dx}=-\frac{2y}{3x^3}}$$

The next problem on his list is finding $\frac{ds}{dt}$ for: $$s^3-t^3=8$$ In this problem our $x$ variable is now a $t$. $$s^3-t^3=8\\ 3s^2\frac{ds}{dt}-3t^2=0\\ 3s^2\frac{ds}{dt}=3t^2\\ \boxed{\frac{ds}{dt}=\frac{t^2}{s^2}}$$

The final is example is finding $\frac{dy}{dt}$ for: $$y^3+\sin(yt)=t^5\\ 3y^2\frac{dy}{dt}+\cos(yt)\cdot\color{red}{[yt]'}=5t^4\\ 3y^2\frac{dy}{dt}+\cos(yt)\cdot\color{red}{[yt]'}=5t^4\\ 3y^2\frac{dy}{dt}+\cos(yt)\cdot[\color{red}{(y)'}(t)+(y)\color{red}{(t)'}]=5t^4\\ 3y^2\frac{dy}{dt}+\cos(yt)\cdot[t\frac{dy}{dt}+y]=5t^4\\ 3y^2\frac{dy}{dt}+t\cos(yt)\frac{dy}{dt}+y\cos(yt)=5t^4\\ 3y^2\frac{dy}{dt}+t\cos(yt)\frac{dy}{dt}=5t^4-y\cos(yt)\\ \frac{dy}{dt}[3y^2+t\cos(yt)]=5t^4-y\cos(yt)\\ \boxed{\frac{dy}{dt}=\frac{5t^4-y\cos(yt)}{3y^2+t\cos(yt)}}$$

(Videos) Compilation of Implicit Differentiation Examples

by PatrickJMT

Here's a video by patrickJMT where he goes over some implicit differentiation examples.

Implicit differentiation is similar to regular differentiation except when taking the derivative of $y$ we tack on a $\frac{dy}{dx}$.

The first example he solves is: $$x^3+y^3=xy$$ When we take the derivative the $x^3$ becomes $3x^2$. When we get to $y^3$ we just take the derivative like we normally do except we add on the $\frac{dy}{dx}$. $$3x^2+3y^2\color{red}{\frac{dy}{dx}}=\color{red}{(x)'}y+x\color{red}{(y)'}\\ 3x^2+3y^2\frac{dy}{dx}=y+x\color{red}{\frac{dy}{dx}}\\ 3y^2\frac{dy}{dx}-x\frac{dy}{dx}=y-3x^2\\ \frac{dy}{dx}(3y^2-x)=y-3x^2\\ \boxed{\frac{dy}{dx}=\frac{y-3x^2}{3y^2-x}}$$

The next example is: $$(x^2y)+(xy^2)=3x\\ [\color{red}{(x^2)'}y+x^2\color{red}{(y)'}]+[\color{red}{(x)'}y^2+x\color{red}{(y^2)'}]=3\\ [2xy+x^2\color{red}{\frac{dy}{dx}}]+[y^2+2xy\color{red}{\frac{dy}{dx}}]=3\\ x^2\frac{dy}{dx}+2xy\frac{dy}{dx}=3-2xy-y^2\\ \frac{dy}{dx}(x^2+2xy)=3-2xy-y^2\\ \boxed{\frac{dy}{dx}=\frac{3-2xy-y^2}{x^2+2xy}}$$

Here's another video with yet more implicit equations to derive. This videos example being: $$\sec(xy)+e^y=3x-4\\ \sec(xy)\tan(xy)\color{red}{[xy]'}+e^y\color{red}{\frac{dy}{dx}}=3\\ \sec(xy)\tan(xy)[\color{red}{(x)'}y+x\color{red}{(y)'}]+e^y\frac{dy}{dx}=3\\ \sec(xy)\tan(xy)(y+x\color{red}{\frac{dy}{dx}})+e^y\frac{dy}{dx}=3\\ y\sec(xy)\tan(xy)+x\sec(xy)\tan(xy)\frac{dy}{dx}+e^y\frac{dy}{dx}=3\\ x\sec(xy)\tan(xy)\frac{dy}{dx}+e^y\frac{dy}{dx}=3-y\sec(xy)\tan(xy)\\ \frac{dy}{dx}[x\sec(xy)\tan(xy)+e^y]=3-y\sec(xy)\tan(xy)\\ \boxed{\frac{dy}{dx}=\frac{3-y\sec(xy)\tan(xy)}{x\sec(xy)\tan(xy)+e^y}}$$

Here's another video with even more examples of implicit differentiation. The first problem he looks at is: $$1+x=\sin(xy^2)\\ 1=\cos(xy^2)\cdot\color{red}{[xy^2]'}\\ 1=\cos(xy^2)[\color{red}{(x)'}y^2+x\color{red}{(y^2)'}]\\ 1=\cos(xy^2)[y^2+x(2y)\color{red}{\frac{dy}{dx}}]\\ 1=y^2\cos(xy^2)+2xy\cos(xy^2)\frac{dy}{dx}\\ 2xy\cos(xy^2)\frac{dy}{dx}=1-y^2\cos(xy^2)\\ \boxed{\frac{dy}{dx}=\frac{1-y^2\cos(xy^2)}{2xy\cos(xy^2)}}$$

The next problem asks for the equation of a tangent line at the point $(1,1)$ on the curve $$x^2+xy+y^2=3$$ So the first step is to find the derivative of this curve. This will tell us the slope for the tangent line. $$2x+[\color{red}{(x)'}y+x\color{red}{(y)'}]+2y\color{red}{\frac{dy}{dx}}=0\\ 2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\\ x\frac{dy}{dx}+2y\frac{dy}{dx}=-2x-y\\ \frac{dy}{dx}(x+2y)=-2x-y\\ \frac{dy}{dx}=\frac{-2x-y}{x+2y}$$ Now we plug in the point $(1,1)$ in order to find the slope at that point. $$m=\frac{-2\color{red}{(1)}-\color{red}{(1)}}{\color{red}{(1)}+2\color{red}{(1)}}\\ m=-\frac33=-1$$ Now that we have a point and a slope we can plug all this into the point-slope formula to get the equation of the tangent line. $$y-y_1=m(x-x_1)\\ y-1=-1(x-1)\\ y=-x+1+1\\ \boxed{y=2-x}$$

(Video) Deriving the Derivative of Inverse Tangent

by PatrickJMT

Here's a good video by patrickJMT showing you how to derive the derivative of inverse tangent. This is helpful because it can be hard to remember all the derivative formulas for inverse trig functions. Furthermore, this is a good procedure to remember because you can use a similar method to derive many derivative formulas, like logarithms.

So the function we have is $$y=\tan^{-1}x$$ and we want to know what $y'$ or, in other words, $\frac{dy}{dx}$. We could just differentiate both sides but this doesn't help since we don't know the derivative for $\tan^{-1}x$. So let's rearrange our function by taking the tangent of both sides. $$\tan y=\tan(\tan^{-1}x)$$ Since a function and its inverse cancel, this now becomes: $$\tan y=x$$ Now we can use implicit differentiation and take the derivative of both sides. $$\begin{align} \frac{d}{dx}(\tan y)&=\frac{d}{dx}(x)\\ \sec^2y\cdot\color{red}{[y]'}&=1\\ \sec^2y\frac{dy}{dx}&=1\\ \frac{dy}{dx}&=\frac{1}{\sec^2y} \end{align}$$ So we want our derivative to be in terms of $x$ but we have a $y$ in there. We do have the expression: $\tan y=x$. So if we can transform $\sec^2y$ somehow into something with $\tan y$. Luckily, we have a nice trig identity to help us out. $$\begin{align} \cos^2y+\sin^2y&=1\\ \frac{\cos^2y}{\cos^2y}+\frac{\sin^2y}{\cos^2y}&=\frac{1}{\cos^2y}\\ 1+\tan^2y&=\sec^2y \end{align}$$ So we can substitute this into our derivative above. $$\begin{align} \frac{dy}{dx}&=\frac{1}{\sec^2y}\\ &=\frac{1}{\color{red}{1+\tan^2y}} \end{align}$$ And since we have our expression $\tan y=x$, we can now express our derivative in terms of $x$. $$\boxed{\frac{dy}{dx}=\frac{1}{1+x^2}}$$