Derivative of a power

Take a look at the following table and try to find the general rule:

$$f (x)$$ $$f'(x)$$
$$x^2$$ $$2x$$
$$x^3$$ $$3x^2$$
$$x^5$$ $$5x^4$$
$$x^{\frac{1}{2}}$$ $$\frac{1}{2}x{-\frac{1}{2}}$$
$$2x^2$$ $$4x$$
$$2x^3$$ $$6x^2$$
$$5x^6$$ $$30x^5$$
$$x^n$$ ?
$$Ax^n$$ ?

Solution:$$$\begin{array}{ll}f (x) =x^n & f '(x) = nx^{n-1} \\ f (x) = A x^n & f '(x) = A nx^{n-1}\end{array}$$$

Now verify the derivatives in the table by trying to identify what is the $$A$$ and what is the $$n$$ in each of the cases.

We have thus obtained a general formula. We need to emphasize that this formula is only applicable when $$n$$ is a rational number. We will see some examples that will show that we need to bear in mind this fact. Note also the following:

  • If we have a function with a square or cubic root or any type of root we can rewrite it with a power, and we can then apply the rule.

  • When $$n=0$$ the derivative is zero, since any number raised to $$0$$ is $$1$$, which is a constant, and therefore the derivative is zero.

Summing up, then, the general formula has been deduced to derive three types of fundamental functions: constant function, linear function and any power. Check it in the following table:

$$f(x)=A$$ $$f'(x)=0$$
$$f(x)=Ax+b$$ $$f'(x)=A$$
$$f(x)=Ax^n$$ $$f'(x)=A\cdot n\cdot x^{n-1}$$

and look at the following examples:

a) $$\begin{array}{ll}{f (x) = 30x + 5} & {f '(x) = 30}\end{array}$$

b)$$\begin{array}{ll} {f(x)=4(x + 1)} & {f '(x) = 4} \end {array}$$

c) $$\begin{array}{ll} {f (x) = 3 (5x+2)} & {f '(x) = 15} \end {array}$$

d) $$\begin{array}{ll} {f (x) = 6 (x^4+5)} & {f '(x) = 6 · 4x^3 = 24x^3} \end {array}$$

e) $$f(x)=\sqrt{x}=x^{\frac{1}{2}}$$ $$f'(x)=\frac{1}{2} x^{\frac{1}{2}-1}=\frac{1}{2}x^{-\frac{1}{2}}=\dfrac{1}{2\sqrt{x}}$$

f) $$f (x) =\sqrt[3]{\sqrt{x^2}}$$ $$f'(x)=\dfrac{2}{3}x^{\frac{2}{3}-1}=\dfrac{2}{3}x^{-\frac{1}{3}}=\dfrac{2}{3\sqrt[3]{x}}$$