Proof tricks

• Inverting the chain rule. Sometimes, you want to find $f’(x)$ but you can find some $g$ such that $(g\circ f)$ is easily differentiable and $g$ is also differentiable. Then, you can let $f’(x) = (g \circ f)’ / g’(f(x))$.

• If you want to show something general across all derivatives, the MVT is a good bet (epsilon-delta definitions may be too tedious)

• Generally, when you’re given a limit of a function, try to use general limit rule sand higher abstractions before you try delta epsilon, as delta-epsilon can get ugly

• If you want to relate a function to its derivative, MVT is a good bet

Basic Properties

What is the derivative? 🐧

We define the derivative of a function to be

A slightly easier definition is

Examples of derivatives from definition 🧸

Derivative of $c$ is 0

• Proof (easy)

  If $f(x) = c$, then $f(x + h) - f(x) = 0$, and we are done.

Derivative of $x^n$ is $nx^{n-1}$

• Proof (binomial expansion)

  We want to compute $(x + h)^n - x^n$, which we can do with binomial expansion. We won’t write it all out here, but it’s $x^n + \binom{n}{1}x^{n-1}h + \binom{n}{2}x^{n-2}h^2 + … h^n - x^n$, and if we divide the whole thing by $h$ in the limit definition, we get $\binom{n}{1}x^{n-1} + …$ and all the other terms have $h$, so as $h → 0$, they go to $0$. We know that $\binom{n}{1} = n$, so the derivative is $n x^{n-1}$ as desired.

Derivative of $\sin(x) = \cos(x)$.

• Proof (trig identities)

  We want to compute $\sin(x + h) - \sin(x)$, which we can do through the angle sum formula: $\sin(x)\cos(h) + \sin(h)\cos(x) - \sin(x)$ and we can group and divide by $h$ (as per the limit setup) to get

  $$\sin(x)(\cos(h) - 1)/h + \cos(x)\sin(h)/h$$

  Now we know that $\lim \sin(h)/h = 1$. This fact can be shown geometrically but we won’t bother here. What remains is to show that $\lim \cos(h) - 1)/h = 0$. To do this, we recognize that $\cos(h) = 1 - 2 \sin^2(h/2)$, which means that we have $(\cos(h) - 1)/h = 2\sin^2(h/2)/ h = \sin(h/2) * \frac{\sin(h/2)}{h/2}$. We know that the second term goes to 1, and the first term goes to 0, so the whole thing goes to zero. As such, only the second term remains: $\lim cos(x) \sin(h)/h = \cos(x)$, as desired.

Derivative of $e^x$ is $e^x$.

• Proof (definition of $e$)

  Right away, we get $(e^{x + h} - e^x)/h = e^x (e^h - 1)/h$. What remains is to compute $\lim (e^h - 1)/h$.

  Let $y = e^h - 1$. Therefore our limit becomes $\lim_{y \rightarrow 0}\frac{y}{ \ln(y + 1)} = \lim \frac{1}{\ln((y+1)^{1/y})}$.

  Now, recall that $e = \lim_{n\rightarrow \infty} ( 1 + 1/n)^n$. So, let $n = 1/y$. Furthermore, as $\ln$ is a continuous function, we can just apply the $\ln$ to the definition. Therefore, we have $\lim \frac{1}{\ln((y+1)^{1/y})} = \frac{1}{\ln(e)}= 1$.

  As such, the whole limit becomes $e^x * 1 = e^x$, which is what we wanted to show.

Differentiability implies continuity 🚀

Theorem: if a function is differentiable at a point $a$, then it must be continuous

This does NOT go the other way! Continuous functions may not be differentiable.

• Proof (definitions) [this does NOT show uniformly continuous. For that, you need to also assume that the derivative is bounded]

  Let’s show the official defiitions. We know that for any $\epsilon$, there exists some $\delta_\epsilon$ such that $|(f(x) - f(x_0))/(x - x_0) - f’(x_0)| < \epsilon$ for $|x - x_0| < \delta_\epsilon$. Let $\epsilon = |f’(x_0)| / 2$.

  Under this setup, we can create the looser bound $-\frac{3}{2}|f’(x_0) \leq (f(x)-f(x_0))/(x-x_0) \leq \frac{3}{2}|f’(x_0)$

  but this means that $|f(x) - f(x_0)| < \frac{3}{2}|x - x_0||f’(x_0)|$. Therefore, we can solve for some new $\delta’_\epsilon = \epsilon/(\frac{3}{2}|f(x_0)|)$.

  As our original string of logic follows from $|x - x_0| < \delta^{deriv}_\epsilon$, we claim that our $\delta_\epsilon = \min(\delta^{deriv}_\epsilon, delta’_\epsilon)$. Because such a $\delta_\epsilon$ exists for all $\epsilon$, we are done.

  This was a more rigorous workup. Here’s an alternative proof:

  We can express $f(x) = (x - a)\frac{f(x)-f(a)}{x - a} + f(a)$, and because $\lim_{x \rightarrow a} (x - a) = 0$ and $\frac{f(x) - f(a)}{x-a}$ is finite as $x \rightarrow a$, we have $\lim f(x) = f(a)$, which means that it is continuous.

Derivative rules 🚀

Scalar rule: $(cf)’(a) = c f’(a)$

• Proof

  This is trivial. Just use the definition of the derivative, factor out a $c$, and you have your answer.

Sum rule: $(f + g)’(a) = f’(a) + g’(a)$

• Proof

  This is also trivial; just use the derivative definitions and split up the numerator into $f(x) - f(a) + g(x) + g(a)$.

Product rule: $(fg)’(a) = g(a)f’(a) + g’(a)f(a)$.

• Proof

  We do the classic add and subtract trick: $f(x+h)g(x+h) - f(x)g(x) = f(x+h)g(x+h) + g(x)f(x+h) - g(x)f(x+h) - f(x)g(x)$.

  From this, we can break up the segments:

  $$f(x+h)\frac{g(x+h) - g(x)}{h} + g(x)\frac{f(x+h)-f(x)}{h}$$

  And because $f$ is continuous, then $\lim_{h\rightarrow 0} f(x+h) = f(x)$. This yields the $f(x)g’(x) + g(x)f’(x)$ form as desired.

Quotient rule: $(f/g)’(a) = (g(a)f’(a) - f(a)g’(a)) / g^2(a)$. [low-dee-high minus high-dee-low is a good way of remembering it].

• Proof

  This one is less beautiful. You basically combine the numerator and denominator

  

  and do a classic add-one-subtract-one trick. Now, when we divide by $x -a$, we can factor like the following:

  

  and then we take the limit, and using continuity properties of $g, f$, we get our formula.

Chain rule: $(g \circ f)’(a) = g’(f(a))f’(a)$.

• Proof (multiply by 1)

  The trick here is to express it as

  $$\frac{g(f(x + h))- g(f(x))}{h} * \frac{f(x + h) - f(x)}{f(x+h) - f(x)}$$

  and we just move the numerators around to get

  $$\frac{g(f(x + h))- g(f(x))}{f(x+h) - f(x)} * \frac{f(x + h) - f(x)}{h}$$

  The second part is just $f’(x)$, but what about the first part? Well, let $\tilde h = f(x + h) - f(x)$ as a variable substitution. Now, we have

  $$\lim_{\tilde h \rightarrow 0} \frac{g(f(x) + \tilde h) - g(f(x))}{\tilde h}$$

  And this is just $g’(f(x))$. We implicitly use the continuity of $f$ here: as $h → 0$, we have that $f(x+h) - f(x) → 0$.

  So…we have $g’(f(x))*f’(x)$, as desired.

Theorems of Derivatives

Increasing, decreasing 🐧

Given some interval $I$

• $f$ strictly increases if $x_1 < x_2 → f(x_1) < f(x_2)$, vice versa for strictly decreases

• $f$ increasing if $x_1 < x_2 \rightarrow f(x_1) \leq f(x_2)$

Extrema zero-derivative theorem 🚀

Theorem: if $f$ is defined on $(a, b)$ and $f(x_0)$ is the maximum or minimum value (not the infimum or supremum), then $f’(x_0) = 0$

• Proof (contradiction)

  INTUITION: At a maximum, if the derivative is positive, we can follow the derivative forward to get a higher number. If the derivative is negative, we can follow it backwards to get a higher number.

  Without loss of generality, assume that $f(x_0)$ is a maximum. Now, suppose that $f’(x_0) > 0$. Then, set $\epsilon = f’(x_0) / 2$, and you get some $\delta$ such that $|x - x_0| < \delta$ implies that $|(f(x) - f(x_0)) / (x - x_0) - f'(x_0)| < f'(x_0)/2$.

  Now, we can use one half of the absolute value definition to get $f(x) - f(x_0) > (x - x_0)f’(x_0)/2$.

  Now, consider the case where $x > x_0$ (i.e. half of the bound $\delta)$. Then, we have $f(x) > f(x_0)$, which contradicts the original definition that $f(x_0)$ is the maximum.

  Similarly, if $f’(x_0) < 0$, we can construct a similar setting where we get a contradiction with $f(x) > f(x_0)$.

  Therefore, we must conclude that $f’(x_0) = 0$

Rolle’s Theorem 🚀

Theorem: if $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$ an dsatisfies $f(a) = f(b)$, then there is at least one $x \in (a, b)$ such that $f’(x) = 0$. In other words, “what goes up must come down.”

• Proof (maxima in a closed interval, extrema zero-derivative)

  Any function in a closed interval has $x_0, x_1$ such that $f(x_0) \leq f(x) \leq f(x_1)$. We proved this previously. If $x_0, x_1 = a, b$ respectively, then $f(x)$ is constant, so $f’(x) = 0$ all throughout.

  Otherwise, some $x_0 \in (a, b)$, and we proved in the extrema zero-derivative theorem that $f’(x_0) = 0$.

Mean Value Theorem 🚀

Theorem: If $f$ is continuous on $[a, b]$ and differentiable on $(a, b)$, then there exists at least one $x \in (a, b)$ such that $f’(x) = (f(b) - f(a))/(b -a)$.

• Proof (construction of a secant line)

  The intuition is that we want to “rotate” the function such that $g(b) = g(a)$ for this new function, and somehow reason from it. Consider the following function:

  $$g(x) = f(x) - (x - a)\frac{f(b) - f(a)}{b - a}$$

  We see that $g(a) = f(a)$ (it’s the untouched function) and $g(b) = f(a)$. The funciton still follows the curve of $f$, but it is “rotated.”

  All elements of $g$ are continuous and differentiable. Because $g(a) = g(b)$, there exists some $c \in (a,b)$ such that $g’(c) = 0$. And if we expand the definition of $g$ again, this fact means that $f’(c) = \frac{f(b) - f(a)}{b-a}$ as desired.

Corollaries of the MVT

Corollary: if $f’(x) = 0$ in an interval, it must be a constant function

• Proof

  Pick any $c \neq d \in [a, b]$. By the mean value theorem, we know that there exists some $e \in (c, d)$ such that $(f(c) - f(d)) / (c - d) = f’(e)$, and because $f’(e) = 0$ and $c \neq d$, we conclude that $f(c) = f(d)$. This is true for any $c, d$ so we conclude that $f(x)$ is constant.

Corollary: if $f’(x) > 0$ in an interval, the function $f$ must be strictly increasing (and vice versa)

• Proof

  Similar setup as above with $c > d$ and $e \in (d, c)$. If this were the case, then we have $(f(c) - f(d))/(c -d) = f’(e) > 0$, which means taht $f(c) - f(d) > 0$ whenever $c > d$, which is what we wanted to show

Corollary: if $|f’(x)| \leq M$ for some finite $M$, then $f$ is uniformly continuous.

• Proof

  Do a similar setup with $c, d$ and $e \in (c, d)$. If this is the case, then we have $|f(c) - f(d)| = |f’(e)||c - d|$. We

  $$|f(c) - f(d)| = |f’(e)||c - d| \leq M|c -d|$$

  And so if we pick some $\epsilon$, we can just let $|c -d| < \epsilon / M = \delta_\epsilon$. NOw, we note that this works for any $c, d$, and the $\delta_\epsilon$ doesn’t depend on $c, d$, so we conclude that it is uniformly continuous.

Corollary: If $f’ = g’$, then $f = g + c$.

• Proof

  If this is not the case, then $f - g$ contradicts the MVT (write it out and you will see why)

Intermediate Value Theorem for Derivatives (bonus) 🚀

Theorem: If $f$ is differentiable in $(a, b )$. If $a < x_1 < x_2 < b$ and $c$ is between $f’(x_1)$ and $f’(x_2)$, then there exists some $x \in (x_1, x_2)$ such that $f(x) = c$.

Derivative of an inverse (bonus)🚀

Theorem: if $f$ is one-to-one, and let $J = f(I)$. If $f$ is differentiable in $I$ and all $f’(x_0) \neq 0$, then

where $y = f(x)$.

Local Extrema 🐧

We define a local extrema of a function $x_0$ to be a point such that there exists some $\delta > 0$ where $f(x_0) > f(x)$ for all $|x - x_0| < \delta$ (or vice versa for a minimum)

We won’t prove this, but to see if $x_0$ is a local maximum, minimum, or neither, we can use a general version of the second derivative test.

1. If $f$ is infinitely differentiable, start with points where $f’(x_0) = 0$. Then, keep on taking derivatives such that $f^{(n)}(x_0) = 0$ and $f^{(n+1)}(x_0)\neq 0$

2. If $n$ is even, then $x_0$ is neither local max or min

3. If $n$ is odd, then if $f^{(n+1)}(x_0)>0$, then $x_0$ is a local minimum. Else, local maximum.

L’Hospital’s Rule 🚀

To find $\lim f(x)/g(x)$ when $f(x_0) = g(x_0) = 0$ or when $f(x_0) = g(x_0) = \infty$, it is sufficient to calculate $\lim f’(x)/g’(x)$.

• Intuition (MVT)

  For sake of simplicity, assume that $f(x_0) = 0$. Then, $f(x) = f(x) - f(x_0)$, and by the MVT, we have that $f(x) - f(x_0) = f’(c)(x - x_0)$ for some $c \in (x, x_0)$. A similar thing can be said for $g$. Therefore, we have $f(x)/g(x) = f’(c_f)(x - x_0) / g’(c_g)(x - x_0) = f’(c_f)/g’(c_g)$. Now, this would have been a sufficient proof if we are able to show that $c_f = c_g$, which at the moment is non-trivial.

From the intuition, we understand that we somehow need to get these points to be the same. To help, we propose the following lemma:

Lemma: if $f, g$ are differentiable in $(a, b)$, then there exists some $c \in (a, b)$ such that

• Proof (Workup from Rolle’s theorem)

  Intuitively, we want to expand our proof of the MVT to a general line. In our previous proof, we “rotated” the $f$ by a secant line. Now, we want to “rotate” it by $g$. Consider the function

  $$h(x) = f(x) - (g(x) - g(a))\frac{f(b) - f(a)}{g(b) - g(a)}$$

  We see that $h(a) = f(a)$ (untouched) and $h(b) = f(a)$. By rolle’s theorem, there must exists some $c \in (a, b)$ such that $h’(c) = 0$. If we use the above definition, this means that $f’(c) - g’(c) \frac{f(b)-f(a)}{g(b)-g(a)} = 0$, and we can easily solve to get the original expression.

Finally, we will introduce a simple concept: the one-sided limit. We say that $\lim_{x →x_0^+}f(x) = L$ if for all $\epsilon > 0$, we have $|f(x) - L| < \epsilon$ for all $x \in (x_0, x_0 + \delta_\epsilon)$. Vice versa for the other side.

A true limit exists if and only if the left and right sided limits exist and are equal.

Now, we are ready for the actual proof.

• Proof (working from $f’/g’$ to the limit of $f/g$.

  Let’s say that $f(x_0) = 0$ for simplicity; we can prove the $\infty$ form in a similar way.

  Suppose that we know that $\lim f’(x) / g’(x) = L$. Let’s show that $\lim f(x)/g(x) = L$.

  Let’s use the initial limit definitions. We know that for all $\epsilon$, we have $|f’(x)/g’(x) - L| < \epsilon$ for all $x \in (x_0 - \delta_\epsilon, x_0 + \delta_\epsilon)$. Now, consider $(a, b) \in (x_0, x_0 + \delta_\epsilon)$ and $(f(b) - f(a))/(g(b) - g(a))$.

  By the our modified MVT lemma, we know that there exists some $c$ such that $f’(c) / g’(c) = (f(b) - f(a))/(g(b) - g(a))$. But we also know that $c \in (a, b)$, which means that $c \in (x_0 - \delta_\epsilon, x_0 + \delta_\epsilon)$.

  Because we know this, we can use our limit definitions and say that

  $$|\frac{f(b) - f(a)}{g(b) - g(a)} - L| = |\frac{f'(c)}{g'(c)} - L| < \epsilon$$

  Hmm! We’re getting somewhere.

  Now, let’s just close the bound and let $a → x_0$ from the positive side. As we are doing this, we not violating any of our above constructions, so the inequality still holds (i.e. there still exists some $c$, so we can use this same reasoning).

  As we have $f$ is continuous, we know that $\lim_{a → x_0} f(a) = f(x_0) = 0$. Same thing with $g$. Therefore, we have

  $$|\frac{f(b)}{g(b)} - L|< \epsilon, b \in (x_0, x_0 + \delta_\epsilon)$$

  Therefore, we conclude by limit definitions that $\lim_{x→x_0^+}\frac{f(x)}{g(x)} = L$. We can apply the same proof method to show that the other side works.

  ☝

  We approach with one-sided limits so that we don’t have to deal with two moving parts.

  Therefore, we conclude that $\lim_{x →x_0}\frac{f(x)}{g(x)} = L$, as desired.

  What we did is show that if the derivative quotient limit is $L$, then the function quotient limit is also $L$.

  As a side note, if $f(x_0) = \infty$, then we need to rewrite the expression as

  $$\frac{f(b)-f(a)}{g(b)-g(a)} = \frac{f(a)}{g(a)}\frac{1 - f(b)/f(a)}{1 - g(b)/g(a)}$$

  and as $a → x_0$, the $f(b)/f(a) → 0$. This is the hand-wavy version of it.