Constant functions and the identity functions are differentiable at all numbers, and their derivatives are given by the following formulas:

• Suppose \(f\) is a constant function; that is, there is some number \(c\) such that \(f(x)=c\). Then \(f'(x)=0\). In other words, \(\dfrac{d}{dx}(c) = 0\).
• Suppose \(f\) is the identity function; that is, \(f(x)=x\). Then \(f'(x)=1\). In other words, \(\dfrac{d}{dx}(x) = 1\).

Suppose that \(f\) and \(g\) are functions and \(c\) is a number.

• If \(f\) and \(g\) are differentiable at a number \(x\), then so is \(f + g\), and

  \((f+g)'(x) = f'(x) + g'(x)\)

  In other words,

  \(\dfrac{d}{dx}(f(x)+g(x)) = \dfrac{d}{dx}(f(x)) + \dfrac{d}{dx}(g(x))\)

• If \(f\) is differentiable at \(x\), then so is \(cf\), and

  \((cf)'(x) = c(f'(x))\)

  In other words,

  \(\dfrac{d}{dx}(cf(x)) = c.\dfrac{d}{dx}(f(x))\)

• If \(f\) and \(g\) are differentiable at a number \(x\), then so is \(f - g\), and

  \((f-g)'(x) = f'(x)-g'(x)\)

  In other words,

  \(\dfrac{d}{dx}(f(x)-g(x)) = \dfrac{d}{dx}(f(x)) - \dfrac{d}{dx}(g(x))\)

If \(f\) and \(g\) are differentiable at \(x\), then so is \(fg\), and

\((fg)'(x) = f(x)g'(x) + g(x)f'(x)\)

In other words,

\(\dfrac{d}{dx}(f(x)g(x)) = f(x)\dfrac{d}{dx}(g(x)) + g(x)\dfrac{d}{dx}(f(x))\)

Suppose \(n\) is a positive integer, and \(f(x) = x^n\). Then \(f\) is differentiable at all numbers, and \(f'(x) = nx^{n-1}\). In other words, \(\dfrac{d}{dx} (x^n) = nx^{n-1}\).

If \(f\) and \(g\) are differentiable at \(x\) and \(g(x) \neq 0\), then \(f/g\) is differentiable at \(x\), and

\((f/g)'(x) = \dfrac{g(x)f'(x) - f(x)g'(x)}{(g(x))^2}\)

In other words,

\(\dfrac{d}{dx}(\dfrac{f(x)}{g(x)}) = \dfrac{g(x)\dfrac{d}{dx}(f(x)) - f(x)\dfrac{d}{dx}(g(x))}{(g(x))^2}\)

Suppose \(n\) is a negative integer, and for all \(x \neq 0, f(x) = x^n\). Then \(f\) is differentiable at all \(x \neq 0\), and \(f'(x) = nx^{n-1}\). In other words, \(\dfrac{d}{dx}x^n = nx^{n-1}\).

Suppose \(n\) is an integer, \(n \geq 2\), and \(f(x) = x^{\dfrac{1}{n}}\). Then \(f'(x) = \dfrac{1}{n}x^{\dfrac{1}{n} - 1}\). If \(n\) is even then the domain of \(f'\) is \((0, \infty)\), and if \(n\) is odd then it is \((-\infty,0) \cup (0, \infty)\).

Ref 1: Definition of Radicals

Ref 2: Domain of Radical Functions

\(\dfrac{d}{dx}(\sin x) = \cos x\)

\(\dfrac{d}{dx}(\cos x) = -\sin x\)

\(\dfrac{d}{dx}(\tan x) = \sec^2 x\)

\(\dfrac{d}{dx}(\cot x) = -\csc^2 x\)

\(\dfrac{d}{dx}(\sec x) = \sec x \tan x\)

\(\dfrac{d}{dx}(\csc x) = -\csc x \cot x\)