Derivative Rules

Differentiability §

A function $f(x)$ is differentiable at $x=a$ (the derivative exists) if the following conditions are met:

1. f(x) is continuous at $x=a$
2. ${\lim }_{x\to a}{f}^{\prime }(x)$ exists (aka ${\lim }_{x\to a^{-}}{f}^{\prime }(x)={\lim }_{x\to a^{+}}{f}^{\prime }(x)$)

A function if non-differentiable at four places:

1. Discontinuities (see continuity)
2. Corners (V shape)
3. Cusps (Fancy V shape)
4. Vertical tangent lines (${\lim }_{x\to a}f(x)$ can exist but it $\to \pm \infty$)

Notation §

You don’t actually have to know the name of either of these I promise. I am just waffling because I think they are cool.

Lagrange’s Notation §

• $f(x)⟶$ a function
• $f^{\prime }(x)⟶$ the first derivative of a function (“f prime”)
• $f^{\prime \prime }(x)⟶$ the second derivative of a function … and so on

If you were to express the derivative of $y=f(x)$ in Lagrange’s notation, you would end up with $y^{\prime }$, but you don’t see that super often.

Leibniz’s Notation §

Not currently used in our class, but I think it is pretty important still. Leibniz’s notation is the type you usually see when you’re talking about derivation. It’s pretty flexible. In this notation, the derivative of $f$ is expressed as $\frac{d}{dx}f(x)$— the derivative of $f(x)$ with respect to $x$. If you have an equation $y=f(x)$, you can write it as $\frac{dy}{dx}$— the derivative of $y$ with respect to $x$.

To do a higher order derivative, you can just add an exponent to the $d$ and what you’re “respecting”— in this case, $x$.

$$y=f(x),\frac{dy}{dx},\frac{d^{2}y}{d{x}^2},\frac{d^{3}y}{d{x}^3}$$

This seems a bit awkward but it actually comes from manipulating the notation itself;

$$\frac{d(\frac{dy}{dx})}{dx}=\left(\frac{d}{dx}\right)\left(\frac{dy}{dx}\right)=\left(\frac{d}{dx}\right)\left(\frac{d}{dx}\right)y={\left(\frac{d}{dx}\right)}^{2}y=\frac{d^{2}y}{d{x}^2}$$

Note that to express a point derivative in Leibniz’s notation, you can write $\frac{dy}{dx}(a)$ or ${\frac{dy}{dx}|}_{x=a}$. The first one is easier. You can also just do something like $\frac{d}{dx}f(a)$ if you’ve got $f(x)$ or another function in that kind of format— “functional notation,” if you’re curious— in there.

If you ever go deeper into calculus, Leibniz’s notation will probably be the notation of choice. It is a little harder to learn but it can be used to express much more meaningful derivation.

Newton’s Notation §

I am 100% sure we will not see this so you can ignore it, but it is nice to know things. Newton’s notation just uses a dot to express derivation. The derivative of $y=f(x)$ becomes $\dot{y}$, the derivative of $f(x)$ is $\dot{f}$. It doesn’t tell you what you are deriving with respect to and is used in physics mostly to represent rate of change.

Newton’s notation gets a little deranged once you go to higher order derivatives. Three or four dots is where I think we should stop, but if we want to go even further we would have to break the LaTeX (the math notation language) and do something like $\stackrel{\bullet \bullet \bullet \bullet \bullet \bullet \bullet \bullet }{y}$… but if you really wanted to express it without torturing yourself you could do, like, $\stackrel{8}{\dot{y}}$, which is the standard way to express higher-order Newtonian derivatives, and in my opinion actually looks kind of cool.

Derivation §

A derivative is a function that models the slope of the tangent line to $f(x)$ at a certain point, ideally for all values $x$ within the domain of $f(x)$. This is the instantaneous rate of change at that certain point.

Using the Limit Definition §

$$f^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}$$

Just the standard limit definition. $h$ is how close you are to the point $x$, and because you want to find the instantaneous slope at $x$ you’d like it to be as small as possible of a difference.

Basic Derivation Rules §

These are the basic derivation rules. There are special snowflake rules for trigonometric, hyperbolic, inverse hyperbolic and other weird functions, but those are their own thing and we can stick with these for now. I recommend puling up this table and the little graphic in a side by side tab with the explanations if you’re just reading through

Rule Name	Function	Derivative	Notes
Constant Rule	$c$	$0$	Where $c$ is a constant
Power Rule	$x^n$	$n{x}^{n-1}$	$n$ is a real number
Constant Multiple	$cf(x)$	$c{f}^{\prime }(x)$	$c$ is a constant
Sum/Difference Rule	$g(x)\pm j(x)$	$g^{\prime }(x)\pm h^{\prime }(x)$
Product Rule	$g(x)h(x)$	$g^{\prime }(x)h(x)+g(x)h^{\prime }(x)$
Quotient Rule	$\frac{g(x)}{h(x)}$	$\frac{g^{\prime }(x)h(x)-g(x{)}^{\prime }h(x)}{[h(x){]}^2}$	$h(x)\ne 0$
Sine Rule	$\sin (x)$	$\cos (x)$
Cosine Rule	$\cos (x)$	$-\sin (x)$
Tangent Rule	$\tan (x)$	${\sec }^2(x)$
Chain Rule	$g(h(x))$	$g^{\prime }(h(x))\cdot h^{\prime }(x)$
Natural Logarithm Rule	$\ln x$	$\frac{1}{x}$

Term Rules (“Basic” Rules) §

These rules usually apply to a “term” in a larger function. Note that every term is a function too! For example, while in $y=x^2+x+5$ we call $5$ a term, it is also technically its own function, $y=5$. Same with $x$ as a term; it is actually saying $y=x$. Our actual function $y$ is just combining the output of all of these equations.

Constant Rule §

The constant rule, applies to a constant term $c$. When taking a derivative of a constant with respect to any variable, it cancels itself out. Therefore, $c^{\prime }=0$. We can see this in practice if we go back to the limit definition (which is the foundation for everything here, by the way, so we’ll go back to it a lot to understand the concepts).

$$\begin{array}{r}{f}^{\prime }(x)=\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ c^{\prime }(x)=\underset{h\to 0}{\lim }\frac{c(x+h)-c(x)}{h}\end{array}$$

For all points on the graph of the constant $c$, we’re going to get… $c$. So, $c(x)=c$. That’s like saying “what is $y=5$ equal to at four?” It’s equal to $y=5$. This also means that $c(x+h)$ is actually equal to $c$, because the value of $c$ does not depend on any other value— aka, $x$ or $h$— because it is a constant.

$$\begin{array}{rl}{c}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{c-c}{h}\\ & =\underset{h\to 0}{\lim }\frac{0}{h}\end{array}$$

Based off of what we know about limits, we can “simplify” a limit before substituting it in.

Waffling §

Think about it in terms of continuity. ${\lim }_{h\to 0}\frac{0}{h}$ can be likened to a “removable discontinuity”, even if it isn’t technically one. When you’re “removing” a removable discontinuity, you re-define the function to “fill in” the spot that is the problem.

$f(x)$ is discontinuous $x=a$. We define a new function $g(x)$ to be equal to $f(x)$, except for at $f(a)$, where we modify the $g(x)$ to instead be continuous at that point.

$$g(x)=\{\begin{array}{ll}f(x)& \text{for }x\ne a\\ L& \text{for }x=a\end{array}$$

$L$ is the value that the new $g(x)$ equals at $x=a$. Because we can choose $L$, we want it to be what makes the function continuous— in the case of a removable discontinuity, $L$ would be equal to ${\lim }_{x\to a}f(x)$, because if it has passed condition $i$ we would know that $f(x)$ exists.

“Proving” the Constant Rule §

But ${\lim }_{h\to 0}\frac{0}{h}$ looks like there’s nothing more to be done when you take a cursory glance. This would mean you would probably go substitute in the zero, and get a yield of $c^{\prime }(x)=\frac{0}{0}=DNE$. But that’s not actually the case.

The truth is, you don’t need any fancy rules to solve ${\lim }_{h\to 0}\frac{0}{h}$. You need to think of it as a pseudo-”removable discontinuity.” When we solve for condition $i$, we tend to use algebraic factoring or other tools we learnt in class to find our solution, but when you come down to the root of that it is just plain logic.

So we can look at the fraction $\frac{0}{h}$ itself, isolated from the limit notation. Zero divided by any value is equal to zero, so we can just plain simplify to get the derivative.

$$\begin{array}{rl}{c}^{\prime }(x)& =\underset{h\to 0}{\lim }0\\ & =0\end{array}$$

Power Rule §

The power rule is applicable to functions $x^n$ where $n$ is a real number. It essentially just states that the derivative of $y=x^n$, $y^{\prime }$, is equivalent to $n{x}^{x-1}$. Multiply the base of the function by the original power, then subtract one from the power.

It’s actually pretty intuitive when you consider the amount of polynomial multiplication you would have to do what you’re presented with monstrously big powers. A lot of people have trouble understanding where exactly this rule comes from.

To demonstrate, we’ll use our good old fashioned limit definition again to find $f^{\prime }(x)$ if $f(x)=x^6+x^5-x^3+9$.

$$\begin{array}{rl}{f}^{\prime }(x)& =\frac{f(x+h)-f(x)}{h}\\ & =\frac{\left[(x^6+x^5-x^3+9)+h\right]-(x^6+x^5-x^3+9)}{h}\\ & =\text{Lol you thought}\end{array}$$

Honestly just looking at that gives me a headache, so there was no way in hell I was going to type it all out. No, there’s a way better way to explain the power rule.

Instead of looking at that long drudgery of an $f(x)$ defined above, we’ll go with a nice $g(x)=x^4$ to demonstrate.

$$\begin{array}{rl}{g}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{g(x+h)-g(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{(x+h{)}^4-x^4}{h}\end{array}$$

We could just multiply this out by hand, but it’s easier to just… apply Pascal’s Triangle to it.

$$\begin{array}{ccccccccc}1& & 4& & 6& & 4& & 1\end{array}$$

$$\begin{array}{rl}(x+h{)}^4& =1x^4+4x^{3}h+6x^2{h}^2+4h^{3}x+1h^4\end{array}$$

So, let’s chuck that back in there:

$$\begin{array}{rl}{g}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{(x^4+4x^{3}h+6x^2{h}^2+4h^{3}x+h^4)-(x^4)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\cancel{x^4}+4x^{3}h+6x^2{h}^2+4h^{3}x+h^4\cancel{-x^4}}{h}\\ & =\underset{h\to 0}{\lim }\frac{4x^{3}h+6x^2{h}^2+4h^{3}x+h^4}{h}\\ & =\underset{h\to 0}{\lim }4x^3+6x^{2}h+4h^{2}x+h^3\\ & =4x^3\end{array}$$

So $g^{\prime }(x)=4x^3$, or $g^{\prime }(x)=4(x^{4-1})$… the power rule. You can kind of see that, well, almost all of that algebra we did was totally senseless. Let’s look at it a more holistic way.

But what actually happens when we find the derivative? Going back to how we solved $g^{\prime }(x)$, we can see that we subtract the expansion of $(x+h{)}^4$ from $x^4$— removing our first value in the expansion of $(x+h{)}^4$, which is $x^4$ itself.

$$\underset{h\to 0}{\lim }\frac{\cancel{x^4}+4x^{3}h+6x^2{h}^2+4h^{3}x+h^4\cancel{-x^4}}{h}$$

The $x^4$ is essentially deleted from existence. We are going back an exponent value, to a function where the base is raised to ${}^3$ instead of ${}^4$. But we are preserving the value of the original exponent— $4$— because it is the coefficient of the second value in the expansion.

Constant Multiple Rule §

The constant multiple rule essentially states that the coefficient of a function is preserved, as long as it is a constant. This often combines with the power rule for functions like $f(x)=4x^3$, where you would multiply the coefficient by $3$ to get $f^{\prime }(x)=12x^2$.

Operand Rules §

Sum/Difference Rule §

Linearity is a key concept in derivation. Linearity essentially states that the derivative of any linear combination of functions— essentially, a sum (or difference, which can be redefined like $x-y=x+(-y)$ and transformed into a sum)— is equal to the sum of the derivatives of those functions.

The sum/difference rule, therefore, just states that the derivative of $g(x)\pm h(x)$ is equal to $g^{\prime }(x)\pm h^{\prime }(x)$.

Product Rule §

Back to the limit definition for this one. We’ll consider $f(x)=g(x)p(x)$

$$\begin{array}{rl}& \\ f^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x-h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\left[g(x+h)p(x+h)\right]-g(x)p(x)}{h}\\ & \\ f^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{g(x+h)\left[p(x+h)-p(x)\right]+p(x+h)\left[g(x+h)-g(x)\right]}{h}\\ f^{\prime }(x)& =g(x)\underset{h\to 0}{\lim }\frac{p(x+h)-p(x)}{h}+p(x)\underset{h\to 0}{\lim }\frac{g(x+h)-g(x)}{h}\\ f^{\prime }(x)& =g(x)p^{\prime }(x)+p(x)g^{\prime }(x)\end{array}$$

Quotient Rule §

And the limit definition for $f(x)=\frac{g(x)}{p(x)}$…

$$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\left[\frac{g(x+h)}{p(x+h)}\right]-\frac{g(x)}{p(x)}}{h}\\ & =\underset{h\to 0}{\lim }\frac{g(x+h)p(x)-g(x)p(x+h)}{h\cdot p(x)\cdot p(x+h)}\\ & =\underset{h\to 0}{\lim }\frac{g(x)\left[p(x+h)-p(x)\right]-p(x)\left[g(x+h)-g(x)\right]}{h\cdot p(x)\cdot p(x+h)}\\ & =\frac{g(x){\lim }_{h\to 0}\frac{p(x+h)-p(x)}{h}-p(x){\lim }_{h\to 0}\frac{g(x+h)-g(x)}{h}}{p(x{)}^2}\\ f^{\prime }(x)& =\frac{g(x)p^{\prime }(x)-p(x)g^{\prime }(x)}{p(x{)}^2}\end{array}$$

Trigonometric Rules §

Sine and cosine rules are given.

Sine-Cosine Swap §

Not a rule, but a quirk that is useful. Every fourth taking of the derivative of sine turns back into $\sin (x)$.

$$\begin{array}{rl}f(x)& =\sin (x)\\ f^{\prime }(x)& =\cos (x)\\ f^{\prime \prime }(x)& =-\sin (x)\\ f^{\prime \prime \prime }(x)& =-\cos (x)\\ f^{\prime \prime \prime \prime }(x)& =\sin (x)\end{array}$$

This can be applied to higher order derivatives as well.

$$\begin{array}{rl}\frac{78}{4}& =19\frac{2}{4}\\ & =\text{We’ve gone through 19 cycles of sine swaps to get here }\\ f^{78}(x)& =-\sin (x)\\ & =\text{Which is the second member of the swap order}\end{array}$$

Tangent Rule §

To derive a rule for $\tan (x)$:

$$\begin{array}{rl}f(x)& =\tan (x)\\ & =\frac{\sin (x)}{\cos (x)}\\ f^{\prime }(x)& =\frac{\cos (x)[\cos (x)]-\sin (x)[-\sin (x)]}{({\cos }^2(x){)}^2}\\ & =\frac{{\sin }^2(x)+{\cos }^2(x)}{{\cos }^2(x)}\\ & ={\sec }^2(x)\end{array}$$

Review trigonometry from Precalc :).

Deriving Further Trigonometric Rules §

All of the basic six trigonometric rules can be derived from just the sine and cosine rules, considering:

$$\begin{array}{rl}\tan (x)& =\frac{\sin (x)}{\cos (x)}\end{array}$$

$$\begin{array}{c}\sec (x)=\frac{1}{\cos (x)},\csc (x)=\frac{1}{\sin (x)},\cot (x)=\frac{1}{\tan (x)}\end{array}$$

For example, to derive a rule for the derivative of $\sec (x)$:

$$\begin{array}{rl}y& =\sec (x)\\ & =\frac{1}{\cos (x)}\\ y^{\prime }& =\frac{(0)[\cos (x)]-(1)[-\sin (x)]}{{\cos }^2(x)}\\ & =\frac{\sin (x)}{{\cos }^2(x)}\\ & =\frac{1}{\cos (x)}\cdot \frac{\sin x}{\cos (x)}\\ & =\sec (x)\tan (x)\end{array}$$

Patterns in Trigonometric Rules §

$$\begin{array}{rlrl}\frac{d}{dx}\left[\sin x\right]& =\cos x& \frac{d}{dx}\left[\csc x\right]& =-\csc x\cot x\\ & & & \\ \frac{d}{dx}\left[\cos x\right]& =-\sin x& \frac{d}{dx}\left[\sec x\right]& =\sec x\tan x\\ & & & \\ \frac{d}{dx}\left[\tan x\right]& ={\sec }^{2}x& \frac{d}{dx}\left[\cot x\right]& =-{\csc }^{2}x\\ & & & \end{array}$$

If you observe the trigonometric derivatives above, you’ll see some patterns. First of all, everything that starts with a c has a derivative that is negative. Both tangent functions $\tan x$ and $\cot x$ are squared.

The most ‘complicated’ are $\csc x$ and $\sec x$, but there’s a pretty simple way to remember them. Both of them repeat themselves and have versions of ‘tan’ in them. The derivative of $\csc$ has $\cot$ in it because it puts $\sin$ on the bottom, while the derivative of $\sec$ has $\tan$ on it because it puts $\cos$ on the bottom. The negative rule still applies here.

Chain Rule §

The chain rule calls on the concept of composite functions to say that the derivative of a composite function is the derivative of its outer function with the inner function substituted in as x, times the derivative of the inner function.

It’s somewhat similar to the product rule; you are taking the value of the derivative of the outer function instead of the original exponent and multiplying it by the inner function instead of the coefficient, then multiplying by the derivative of the inner function instead of changing the power. So not really like the product rule at all, but that is the first thing that comes to mind for me, at least…

More Rules §

Exponential Rules §

This is applicable when you’re given something like $e^{x+1}$ or $a^{2x}$ or whatever. The basic “exponential rule” is to multiply the function times the natural logarithm of its base times the derivative of the exponent. For any function $a^u$:

$$\frac{d}{dx}\left[a^u\right]=a^u\cdot \left[(\ln a)\frac{du}{dx}\right]$$

In practice, this separates into four “rules”, where $a$ represents any base except $e$ and $u$ represents any power raised to besides straight up $x$:

$$\begin{array}{rlrl}\frac{d}{dx}\left[e^x\right]& =e^x& \frac{d}{dx}\left[e^u\right]& =e^u\cdot \frac{du}{dx}\\ & & & \\ \frac{d}{dx}\left[a^x\right]& =a^x\cdot \ln a& \frac{d}{dx}\left[a^u\right]& =a^u\cdot \ln a\cdot \frac{du}{dx}\end{array}$$

Natural Logarithm Rule §

To differentiate natural logarithms, you can put the derivative of the inside over the inside.

$$\frac{d}{dx}\left[x^2+3\right]=\frac{2x}{x^2}=\frac{2}{x}$$

Further Derivative Reading §

1. Implicit Derivation