Implicit Differentiation
[greybox2] Last updated March 13th @ 10:30p: I rewrote the motivation section in hope that the new version is more clear. [/greybox2]
Today’s plan
[greybox2]
- Midterm return and sanity checking (see below).
- How to implicit differentiate
- iClicker
- Problem 1a-1c in Section 3.6. [/greybox2]
Midterm return and sanity checking
[greybox2] I will return your midterms at the beginning of class. Please check the following when you get yours back:
- The score you got on each problem matches the front on the exam.
- The score on the front of your exam was totaled correctly.
If you believe your exam has been graded incorrectly, please bring your exam to me so I can determine what to do next. [/greybox2]
Motivation
Suppose that you have some kind of complicated equation like
\[\begin{align*} x^2y^5 - 4x^4 y^2 + 2xy - 4x + 5y = 3. \end{align*}\]Furthermore, imagine that it is super important to compute $\dfrac{dy}{dx}$. One way to do this is to try to take the above and solve for $y$:
\[\begin{align*} y = \underbrace{(\text{formula after shit-ton of algebra})}_{f(x)}. \end{align*}\]From here, all we would have to do is use our derivative rules to take the derivative of whatever $f(x)$ is with respect to $x$. There is a big issue here, though: the equation you have to solve for $y$ is stupidly hard to solve. Implicit differentiation allows us to sidestep this issue by skipping the step of solving for $y$. The idea is:
- Pretend that it’s actually possible to solve for $y$. This gives you $y = f(x)$. You don’t actually know what $f(x)$ is, but you assume that it is possible to find it.
- (Optional) The fact that this is actually true is a major theorem of mathematics known as the Implicit Function Theorem. You are NOT expected to know this theorem – this is just something I wanted to mention for fun in case anyone is interested. :)
- The derivative $\dfrac{dy}{dx}$ that you obtain through implicit differentiation is secretly the derivative of $f(x)$ with respect to $x$.
How to implicit differentiate
[greybox] Section 3.6, Problem 1a. Find $\dfrac{dy}{dx}$, using implicit differentation, given
\[\begin{align*} 2xy = 3x^2 + 4y^2. \end{align*}\][/greybox]
[solution] Since $y$ depends on $x$ and we want to find the derivative of $y$ with respect to $x$, we use the three following steps:
(1) Take $\dfrac{d}{dx}$ of both sides.
We take the equation and slap a $\frac{d}{dx}$ on both sides:
\[\begin{align*} \frac{d}{dx} (2xy) = \frac{d}{dx}(3x^2 + 4y^2). \end{align*}\](2) Differentiate each term with respect to $x$, being careful to include $\dfrac{dy}{dx}$ whenever we differentiate a term involving $y$.
Let’s look at the stuff on the left first: $\dfrac{d}{dx}(2xy)$. Since $y$ depends on $x$, we think of $y$ as a function of $x$ which means we use product rule to obtain
\[\begin{align*} \frac{d}{dx}(2xy) = 2y + 2x\frac{dy}{dx}. \end{align*}\]Now let’s look at the stuff on the right: $\dfrac{d}{dx}(3x^2 + 4y^2)$. We can use stuff we’ve learned in the past to differentiate $3x^2$, the trickier part is $4y^2$. Since $y$ depends on $x$, we think of $y$ as a function of $x$. So its derivative is obtained by chain rule. Putting all that togther, we get
\[\begin{align*} \dfrac{d}{dx}(3x^2 + 4y^2) = 6x + 8y\frac{dy}{dx}. \end{align*}\]Thus, our final equation for this step is
\[\begin{align*} 2y + 2x\frac{dy}{dx} = 6x + 8y\frac{dy}{dx}. \end{align*}\]3. Rearranging for $\dfrac{dy}{dx}$.
Bringing all the terms involving $\dfrac{dy}{dx}$ on the left and all the others on the right, we obtain
\[\begin{align*} 2x\frac{dy}{dx} - 8y\frac{dy}{dx} = 6x - 2y. \end{align*}\]We can then factor $\dfrac{dy}{dx}$ through the left side to obtain
\[\begin{align*} (2x - 8y)\dfrac{dy}{dx} = 6x - 2y. \end{align*}\]And then we divide both sides by $(2x - 8y)$ to get $\dfrac{dy}{dx}$ by itself. This gives us
\[\begin{align*} \frac{dy}{dx} = \frac{6x - 2y}{2x - 8y}. \end{align*}\][/solution]
Lecture Problems
[greybox] Section 3.6, Problem 1a. Find $\dfrac{dy}{dx}$, using implicit differentation, given
\[\begin{align*} 2xy = 3x^2 + 4y^2. \end{align*}\]Obviously we did the problem above and in lecture, but try to reattempt it on your own without looking at how I did it. [/greybox]
[solution]
\[\begin{align*} \frac{dy}{dx} = \frac{6x - 2y}{2x - 8y}. \end{align*}\][/solution]
[greybox] Section 3.6, Problem 1b. Find $\dfrac{dy}{dx}$, using implicit differentation, given
\[\begin{align*} \frac{x}{y} = 5x^2 + 2y \end{align*}\][/greybox]
[solution]
\[\begin{align*} \frac{dy}{dx} = \frac{y - 10xy^2}{2y^2 + x}. \end{align*}\][/solution]
[greybox] Section 3.6, Problem 1c. Find $\dfrac{dy}{dx}$, using implicit differentation, given
\[\begin{align*} \sqrt{xy} = 3 + y^3. \end{align*}\][/greybox]
[hint] Rewrite $\sqrt{xy}$ using exponent laws. [/hint]
[solution]
\[\begin{align*} \frac{dy}{dx} = \frac{-y}{x - 6y^2\sqrt{xy}}. \end{align*}\][/solution]