If \(g\) is differentiable at \(x\) and \(f\) is differentiable at \(g(x)\), then \(f \circ g\) is differentiable at \(x\), and

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

In other words,

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

We can streamline our use of the chain rule by noticing the relationship between the following two equations:

\(\dfrac{d}{dx}(f(x)) = f'(x)\)

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

The first equations says that the derivation of \(f(x)\) is \(f'(x)\). Replacing \(x\) with the expression \(g(x)\) on both sides, you might expect that the derivative of \(f(g(x))\) would be \(f'(g(x))\). The chain rule says that this is part of the answer, but it's not the whole thing. To get the correct answer you need to multiply by the derivative of \(g(x)\)

Suppose \(f(x) = x^r\), where \(r\) is a nonzero rational number. Then \(f'(x) = rx^{r-1}\).

Since \(r\) is rational and \(r \neq 0\), we can write \(r=m/n\), where \(n\) is a positive integer, \(m\) is a nonzero integer, and the fraction \(m/n\) is reducted to lowest terms. If \(n\) is even, then the domain of \(f'\) is \((0,\infty)\). If \(n\) is odd and \(r<1\), then the domain of \(f'\) is \((-\infty,0) \cup (0,\infty)\). And if \(n\) is odd and \(r \geq 1\) then the domain of \(f'\) is \((-\infty,\infty)\)

If \(f\) is a function, then its derivative \(f'\) is also a function. The derivative of \(f'\) is written as \(f''\); it's called the second derivative of \(f\). If we differentiate \(f\) \(n\) times, then we get the \(n\) th derivative of \(f\), which is denoted \(f^{(n)}\). These are called *higher-order derivatives$ of \(f\).