Differentiation rules

Common answer:
Given $$f(x) = ax^n,$$ where $a$ and $n$ are constants, the derivative of $f$ is $$f'(x) = anx^{n-1},$$ where $f'(x)$ is the derivative of the function $f$ with respect to $x$.

Equivalently, we can write the derivative of $f(x) = ax^n$ as $$\dfrac{df(x)}{dx} = anx^{n-1}.$$

The actually correct answer:
The above answer is for $n \neq 0$. So, we need to tell our reader what happens if $n = 0$.

In the case where $n = 0$, the function $f$ simplifies to $$f(x) = a,$$ which is a constant. And the derivative of a constant is zero, $$f'(x) = 0.$$

So, the correct answer is actually $$f (x) = \left\lbrace \begin{matrix} anx^{n-1}, & \text{for } n \neq 0 \\ 0, & \text{for } n = 0. \end{matrix} \right.$$ But, most people prefer to ignore the writing out the exception and would accept the simpler (but not fully correct) answer.

The derivative of a sum of functions is equal to the sum of the derivatives of the functions. That is, for $$h(x) = f(x) + g(x),$$ its derivative is $$h'(x) = f'(x) + g'(x).$$ Equivalently, we can write it as $$\dfrac{dh(x)}{dx} = \dfrac{df(x)}{dx} + \dfrac{dg(x)}{dx} .$$

Given a function that is the product of two functions, $$h(x) = f(x) \, g(x),$$ its derivative is $$h'(x) = f'(x) \, g(x) + f(x) \, g'(x).$$ Equivalently, we can write it as $$\dfrac{dh(x)}{dx} = \left(\dfrac{df(x)}{dx}\right)\, g(x) + f(x) \, \left( \dfrac{dg(x)}{dx}\right) .$$

Let the function $h$ be a composition of two functions, $$h(x) = (f \circ g)(x) = f (g (x)).$$ Its derivative is then $$h'(x) = f'(g(x))g'(x).$$ Equivalently, we can write it as $$\dfrac{dh(x)}{dx} = \dfrac{df(g(x))}{dg(x)}\dfrac{dg(x)}{dx}.$$

The derivative of $f(x) = \sin(x)$ is $$f'(x) = \cos (x).$$ Equivalently, we can write it as $$\dfrac{df(x)}{dx} = \cos (x).$$

The derivative of $f(x) = \sin(x)$ is $$f'(x) = -\sin (x).$$ Equivalently, we can write it as $$\dfrac{df(x)}{dx} = -\sin (x).$$

The derivative of $f(x) = e^{x}$ is $$f'(x) = e^{x}.$$ The exponential function, $\exp(x) = e^x$, is a very special in that its derivative is itself.

The derivative of $f(x) = \ln (x)$ is $$f'(x) = 1/x.$$ Equivalently, it can be written as $$\dfrac{df(x)}{dx} = \dfrac{1}{x}.$$

Since we can rewrite $h(x)$ as \begin{eqnarray} h(x) &=& f(x) - g(x) \\ \ &=& f(x) + (-g(x)), \end{eqnarray} we can apply the Sum rule to get \begin{eqnarray} h'(x) &=& f'(x) + (-g'(x)) \\ \ &=& f'(x) - g'(x). \end{eqnarray} Equivalently, we can write $h'(x)$ as $$\dfrac{dh(x)}{dx} = \dfrac{df(x)}{dx} - \dfrac{dg(x)}{dx}.$$

Since we know that $1/x$ can be rewritten as $\dfrac{1}{x} = x^{-1}$ so that we may rewrite $f(x)$ as $$f(x) = x^{-1}.$$ We then apply the Power rule to get \begin{eqnarray} f'(x) &=& -x^{-2} \\ \ &=& \dfrac{-1}{x^2}. \end{eqnarray}

Short solution:
First, we rewrite $f$ as $$ f(x) = [ g(x) ]^{-1}.$$ This form should remind you of the Power rule, which states that $$(ax^n)' = anx^{n-1} \text{ for } n \neq 0. $$ Then we apply the chain rule to $f$ to get \begin{eqnarray} f'(x) &=& -(g(x))^{-2} g'(x) \\ \ &=& \dfrac{-g(x)}{(g(x))^2}. \end{eqnarray}

Long solution:
Notice that we may write $f$ as a composition of two functions. That is, we may rewrite $f(x)$ as $$f(x) = h(g(x)),$$ where $$h(y) = \dfrac{1}{y}.$$ From the previous problem, we know that the derivative of $h$ with respect to $y$ is given by $$h'(y) = \dfrac{-1}{y^2}.$$ Then, we may apply the Chain rule to get \begin{eqnarray} f'(x) &=& h'(g(x))\, g'(x) \\ \ &=& \dfrac{-1}{(g(x))^2} \, g'(x) \\ \ &=& \dfrac{-g'(x)}{(g(x))^2}. \end{eqnarray}
Once we are familiar enough with the basic differentiation rules, we can simplify the derivation above as \begin{eqnarray} f'(x) &=& \left(g(x)^{-1}\right)' \\ \ &=& -g(x)^{-2}\, g'(x) \\ \ &=& \dfrac{-g'(x)}{(g(x))^2}. \end{eqnarray}

We may rewrite $h(x)$ as $$h(x) = f(x)\, ({g(x)})^{-1}.$$ We then apply the Product rule to get \begin{eqnarray} h'(x) &=& f'(x)\, ({g(x)})^{-1} + f(x)\, \lbrace ({g(x)})^{-1} \rbrace' \\ \ &=& f'(x)\, ({g(x)})^{-1} + f(x)\, \lbrace-(g(x))^{-2}\rbrace \\ \ &=& (g(x))^{-2} \lbrace f'(x)\, g(x) - f(x)\, g'(x)\rbrace \\ \ &=& \dfrac{f'(x)\, g(x) - f(x)\, g'(x)}{(g(x))^2}. \end{eqnarray} Notice that we actually made use of the "Reciprocal rule": \begin{eqnarray} \left(\dfrac{1}{g(x)}\right)' &=& \left(g(x)^{-1}\right)' \\ \ &=& -g(x)^{-2}\, g'(x) \\ \ &=& \dfrac{-g'(x)}{(g(x))^2}. \end{eqnarray}
Note: Despite how wide-spread this rule is taught, it is often easier and less error-prone to use the Product rule instead. This is due to two reasons: 1. It is easy to forget which term gets the negative sign. 2. The step where we collect the terms over a common denominator is often not useful and a common source of simple algebraic errors.