Every calculus student has to study the definitions of limits and derivatives (future post), the properties of derivatives of general functions (future post), the derivatives of specific classes of functions, and the applications of derivatives (future post). There will be a future post on how to use these four posts when studying for a test. \(
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Power Rule-Based

These are all of the functions whose derivatives come from the power rule. To understand these, you'll need to have a good handle on your exponent rules<!TODO>, specifically what the definitions of negative and fractional exponents are. We'll also be relying a lot on derivatives distributing over summation.


A polynomial is a sum of terms, each of the form \(c_nx^n\), where \(c_n\) is a constant. To take the derivative of a polynomial, distribute the derivative of the sum, and then apply the power rule to each term, getting results of the form \(nc_nx^{n-1}\). I usually write this out of my formula sheet as the power rule:
\[ \derivop{x^n} = nx^{n-1}dx \]
and leave out the distributing over summation and pulling out coefficients. If you prefer to have that in, you might write it in one of the following ways, for a polynomial of degree \(k\):
\derivop{c_k x^k + \cdots + c_2 x^2 + c_1 x + c_0} \\
= (kc_k x^{k-1} + \cdots + 2c_2 x + c_1)dx \\
\derivop{\sum_{n=0}^k c_nx^n} = \sum_{n=1}^{k} c_nx^{n-1}dx = \sum_{n=0}^{k-1} c_{n+1}x^ndx

Some students prefer to include an example instead of a formula:
\text{If:}&& f(x) &= x^7 - 12x + 4 \\
\text{Then:}&& f'(x)
&= (x^7 - 12x + 4)'
\\ &&&= (x^7)' + (- 12x)' + (4)'
\\ &&&= 7x^6 - 12 + 0

For the other sections, I'll present formulas in my format, assuming you can expand it out or create an illustrative example on your own. My format will be explained more thoroughly in a future post.<!TODO>

Note on Linear Functions.
The derivative of every line is its slope, since if \(f(x) = mx + b\), then \(f'(x) = m\). This is good, since the derivative at a point should be the slope of the line tangent to the function at that point, and we would expect the line tangent to a line to be the same line.

Note on Higher-Order Derivatives.
Since the power rule lowers the powers of \(x\) by \(1\) in the exponent until it's \(0\), the exponents stay nonnegative integers, and the derivatives of polynomials are still polynomials. Further, every polynomial of degree \(k\) has a constant \(k\)th derivative and a zero \((k+1)\)th derivative.


Division by \(x\) can be treated as multiplication by its reciprocal \(x^{-1}\) so that the power rule may be applied. In symbols:
\[\derivop{\dfrac{1}{x^n}} = \derivop{x^{-n}} = -nx^{-n-1}\deriv{x}\]
Since this works for any function in place of \(x\), this acts as a replacement for quotient rule, if you prefer it.

Exercise. Take the derivatives of \((x^n)^{-1}\) and \((x^{-1})^n\) using chain rule, and make sure you get the same result as above.


Taking the \(m\)th root can be treated as raising to the \((1/m)\)th power so that the power rule may be applied. In symbols:
\[ \derivop{\sqrt[m]{x^n}} = \derivop{x^{n/m}} = \frac{n}{m}x^{n/m-1}\deriv{x} = \frac{n}{m}\sqrt[m]{x^{n-m}}\deriv{x} \]
where the last part is only useful in understanding how when you start with \(\sqrt[3]{x^{10}}\), its derivative is \(\frac{10}{3}\sqrt[3]{x^7}\), whose derivative is \(\frac{70}{9}\sqrt[3]{x^4}\), and so on. Usually, though, I prefer avoiding radical notation entirely, and leaving answers with fractional exponents instead.

Note on Absolute Value.<!TODO>
Since the square root function always returns the positive square root, you can define absolute value as \(\abs{x} = \sqrt{x^2}\). That means you can find the derivative of the absolute value, if you are careful:
\derivop{\abs{x}} &= \derivop{\sqrt{x^2}}
\\ &= \derivop{(x^2)^{1/2}}
\\ &= \frac12(x^2)^{-1/2}\derivop{x^2}
\\ &= \frac12(x^2)^{-1/2}(2x)\deriv{x}
\\ &= \frac{x}{\sqrt{x^2}}\deriv{x}
\\ &= \frac{x}{\abs{x}}\deriv{x} = \frac{\abs{x}}{x}\deriv{x}
where the last line is a property of absolute value you can prove separately by multiplying numerator and denominator by \(\abs{x}\) and using \(\abs{x}^2=x^2\). The thing to be careful about is to not oversimplify with \((x^2)^{-1/2}=x^{-1}\), or similar mistakes, along the way. I like to use this example to highlight how careful you must be when making cancellations in exponents — remember that when you cancel an even number, you pick up an absolute value.<!TODO>


Most students memorize the formula
\[ \derivop{e^x}=e^x\deriv{x} \]
If you want a little more reason for that, I'll eventually write a post about the definition of \(e\), which is intrinsically tied to \(e^x\) being its own derivative.

When dealing with an exponential function of a different base, say the constant \(k\), you will need to use the definition of the logarithm to turn it into base \(e\) and use chain rule:
\[ k^x = \left(e^{\ln k}\right)^x = e^{(\ln k)x} \\
\derivop{e^{(\ln k)x}} = e^{(\ln k)x}\cdot\derivop{(\ln k)x} = k^x\cdot(\ln k)\deriv{x} \]
Exercise. Instead use \(k^x = \left(e^x\right)^{\ln k}\) and take the derivative using chain rule and power rule. You get the same answer, in about as many steps.

Now, what about logarithms? Since they are the inverse function of exponentials, we can use that definition to implicitly differentiate and find:
\derivop{\ln x} = \frac{\derivxone}{x}

When you're worried about non-natural logarithms, say base \(k\), you can follow the implicit differentiation steps with \(k^y\) instead of \(e^y\), using its derivative of \((\ln k)k^x\).

Be careful with the derivative of a power where both the base and exponent are functions, say \(u^v\). If you were to assume you could use the power rule or exponential rule, you'd get one of these very different results:
\derivop{u^v} &\wrong{\ne} v u^{v-1}\deriv{u} &&\wrong{\text{power rule}}\\
\derivop{u^v} &\wrong{\ne} (\ln u) u^v\deriv{v} &&\wrong{\text{exponent rule}}
The correct result is actually the sum of these:
\derivop{u^v} &= \derivop{e^{(\ln u)v}}
\\ &= e^{(\ln u)v}\derivop{(\ln u)v}
\\ &= u^v\left(\frac{v}{u}\deriv{u} + (\ln u)\deriv{v}\right)
\\ &= vu^{v-1}\deriv{u} + (\ln u)u^v\deriv{v}
If you find that an odd coincidence, it actually holds in general as the Multivariable Chain Rule, a result you won't see until Calculus III.


From the definitions of the trig functions, all of the functions come from sine and cosine, so once we know their derivatives, we'll know them all. I'll discuss how to find the derivatives of sine and cosine in a future post, but most students just memorize them:
\derivop{\sin x} &= \cos x \\
\derivop{\cos x} &= -\sin x
When I used to get the minus sign mixed up, I would remember what the graphs of sine and cosine look like. For instance, since sine is increasing around \(0\), its derivative must be positive, and since cosine is decreasing right after \(0\), it must become negative. Also, after we find the derivatives of all the trig functions, there we'll notice that the co-functions are exactly the ones with negative signs in the derivative.

To get the derivatives of \(\tan x\) and \(\cot x\), use their definitions and apply quotient rule (or use implicit differentiation). After applying a Pythagorean identity you have what you want:
\derivop{\tan x} &= \sec^2 x \\
\derivop{\cot x} &= -\csc^2 x

To get the derivatives of \(\sec x\) and \(\csc x\), use their definitions and apply quotient rule (or chain rule with the reciprocal function). You then have to split it up into two factors, one that's secant (or cosecant) and another that's tangent (or cotangent). The result:
\[ \begin{aligned}
\derivop{\sec x} &= \sec x\tan x \\
\derivop{\csc x} &= -\csc x\cot x

Now that we have all six derivatives, there is a pattern here. If we sort the functions into \(\DeclareMathOperator{\trig}{trig}\DeclareMathOperator{\arctrig}{arctrig}\trig\) and \(\DeclareMathOperator{\cotrig}{cotrig}\DeclareMathOperator{\arccotrig}{arccotrig}\cotrig\) functions, the \(\trig\) derivatives have no negative signs, and the derivative of \(\cotrig x\) is the negative of the derivative of \(\trig x\) with all the co's swapped. To make this clear, I always wrote the derivatives on my formula sheet in two rows (as they are) and three columns, all next to each other. Then I only had to memorize (or derive) the \(\trig\) derivatives in the first row to get all six.

Take the derivatives of both sides of the identities, particularly the Pythagorean and double angle formulas, and verify that the derivatives are equal.

As stated in the note on the co-function identities, it happens to be true that \(\cotrig x=\trig\left(\frac{\pi}{2}-x\right)\) for each trigonometric function. Applying chain rule to that explains the behavior of the six normal derivatives, as well as an important behavior for the inverse functions. Since the definition of an inverse function swaps<!TODO> the \((x,y)\)-values for \((y,x)\), the fact that \(\cotrig x\) is a horizontal shift and reflection away<!TODO> from \(\trig x\) means that their inverses will be a vertical shift and reflection apart. (In symbols: \(\arccotrig x = \frac{\pi}{2}-\arctrig x\).) Then the derivative of \(\arccotrig x\) is just the negative of the derivative of \(\arctrig x\) for every trig function.

So, the recommendation is: find/memorize the derivatives of \(\arcsin x\), \(\arctan x\), and \(\DeclareMathOperator{\arcsec}{arcsec}\arcsec x\), and the derivatives for the "co" versions will just be their negatives. To find the inverses' derivatives, start with their inverse function definition<!TODO> and use implicit differentiation. You get:
\derivop{\arcsin x} = \frac{\derivxone}{\sqrt{1-x^2}} = -\derivop{\arccos x} \\
\derivop{\arctan x} = \frac{\derivxone}{1+x^2} = -\derivop{\arccot x} \\
\derivop{\arcsec x} = \frac{\derivxone}{\abs{x}\sqrt{x^2-1}} = -\derivop{\arccsc x}

Though you may need to memorize the derivatives of the arc functions for a test in Calculus I, I find them not very useful. In particular, when they start asking you to have them on memory backwards for integration, you don't even need them; you can just learn to use trig substitution<!TODO>, instead.