Lesson 2.3: Differentiating Combinations of Functions

By this point, we’ve covered the differentiation of a few basic functions: monomials and the two elementary trig functions. Obviously, this is woefully inadequate for the majority of functions. There are four “simple” ways to combine two functions $f$ and $g$:

  • $S=f+g$, the sum
  • $D=f-g$, the difference
  • $P=fg$, the product
  • $Q=\frac{f}{g}$, the quotient

The functions $f$ and $g$ can be any functions (though we’ll limit ourselves to constant, monomial, trigonometric, exponential, and logarithmic) of one variable. Additionally, we’ve already dealt with showing that $\frac{d}{dx}\left({cf(x)}\right)=c\frac{df(x)}{dx}$ for a constant $c$; this is called the Constant Multiple Rule.

We’ll first deal with the sum function $S$. Consider $f(x)=2x$ and $g(x)=3x$. It’s clear not only that $f'(x)=2$ and $g'(x)=3$, but also that $S(x)=2x+3x=5x$ and thus $S'(x)=5$

The astute ones among you will notice that $S'=f'+g'$. Coincidence? No. It’s a trivial proof (and a good one for you to do as an exercise) using difference quotients that $\frac{d}{dx}\left({f(x)+g(x)}\right) = \frac{df(x)}{dx}+\frac{dg(x)}{dx}$ and this is known as the Sum Rule.

The rule for the difference function $D(x)=f(x)-g(x)$ is very similar, and naturally called the Difference Rule: $\frac{d}{dx}\left({f(x)-g(x)}\right) = \frac{df(x)}{dx}-\frac{dg(x)}{dx}$ and can be proven using the proof of the Sum Rule in conjunction with the Constant Multiple Rule with $c=-1$.

If you’re as smart as Leibniz was—and that’s quite smart indeed—you’ll be thinking that $\frac{d}{dx}\left({{f(x)\cdot g(x}}\right) = f'(x)\cdot g'(x)$. And if you think so, you’re just as wrong as he was. He did fall into that trap briefly, but then saw what’s wrong with that picture: if that’s true, then the derivative of anything can be shown to be zero. Consider $a(x)=x$, which we can just as easily write as $a(x)=1x$ Now we’ve really got two functions multipled together: $f(x)=1$ and $g(x)=x$. Therefore $f'(x)=0$ and $g'(x)=1$, giving $f'(x)g'(x) = 0\cdot1 = 0$ but we already know that $a'(x)=1$. There’s your disproof by contradiction.

It’s not a particularly easy derivation (though not that horrible) to find the Product Rule for $P(x)=f(x)\cdot g(x)$ I’m not aware of any intuitive way to think of it, but the rule is $\frac{dP(x)}{dx} = \frac{d}{dx}\left({f(x)\cdot g(x)}\right) = f(x)\frac{dg(x)}{dx} + g(x)\frac{df(x)}{dx}$ and the proof is at the link just above. You’ll definitely need to memorize that, but no worries about the order because it’s a sum.

Similarly, it’s incorrect to think that $\frac{d}{dx}\left({\frac{f(x)}{g(x)}}\right) = \frac{f'(x)}{g'(x)}$. (Convince yourself that this would imply that the derivative of anything is undefined.) It’s an even uglier derivation than the Product Rule’s, but the Quotient Rule states that $\frac{dQ(x)}{dx} = \frac{d}{dx}\left({\!\frac{f(x)}{g(x)}\!}\right) = \frac{f'(x)\cdot g(x)—f(x)\cdot g'(x)}{g(x)^2}$

It’s annoying to have to memorize the Quotient Rule, but you certainly need to as it’s frequently useful and a bitch to rediscover every time. Fortunately, generations of students have developed a totally uninspired mnemonic: low dee-high minus high dee-low; draw a line and square the low. “Low” refers to the bottom fraction, $g(x)$, “high” to $f(x)$, “dee-” to differentiating, and the last line refers to making it a fraction. It’s a pretty dumb mnemonic, but I can’t think of any better one. (If you can, you get an A for the day. Seriously, though, let me know).

For your convenience, here’s a summary:

  • the Constant Multiple Rule: $(cf)' = c(f')$
  • the Sum Rule: $(f+g)' = f'+g'$
  • the Difference Rule: $(f-g)' = f'-g'$
  • the Product Rule: $(fg)' = f'g+fg'$
  • the Quotient Rule: $\left({\!\frac{f}{g}\!}\right) = \frac{f'g-fg'}{g^2}$

Now thanks to the Quotient Rule—note that this is the only time you’ll ever be thankful for it—we can find derivatives for the other trigonometric functions:

  • $\frac{d}{dx}\tan x = \frac{d}{dx}\left({\!\frac{\sin x}{\cos x}\!}\right) = \sec^2 x$
  • $\frac{d}{dx}\csc x = \frac{d}{dx}\left({\!\frac{1}{\sin x}\!}\right) = -\csc x \cot x$
  • $\frac{d}{dx}\sec x = \frac{d}{dx}\left({\!\frac{1}{\cos x}\!}\right) = \tan x \sec x$
  • $\frac{d}{dx}\cot x = \frac{d}{dx}\left({\!\frac{\cos x}{\sin x}\!}\right) = -\csc^2 x$

I’d advise against memorizing the derivatives of the cosecant, secant, and cotangent functions; just memorize those for sine, cosine, and tangent and derive the others on the rare occasion you need them.

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