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