Derivatives of polynomials

We know that a derivative of a function $f(x)$ is a limit of the ratio of the function increment to the argument increment when the argument increment tends to $0$:

$$f'(x) = \dfrac{d}{dx}f(x)=\lim\limits_{\triangle x \to 0} \dfrac{f(x+\triangle x)-f(x)}{\triangle x}$$Finding the derivative using this definition is laborious. There are special formulas for finding derivatives of various functions. Let's discuss these rules in this topic.

The power rule

Consider the function $f(x) = 3$. We notice that $$f(x+\triangle x) = 3$$and

$$f(x) = 3$$

because, no matter what point on the function we are talking about, it's equal to $3$. So,

$$f(x+\triangle x)- f(x) = 3-3= 0$$Therefore, we can conclude that a derivative of our function is equal to 0:

$$f'(x) =\lim\limits_{\triangle x \to 0} \dfrac{f(x+\triangle x)-f(x)}{\triangle x} = \lim\limits_{\triangle x \to 0} \dfrac{0}{\triangle x}=0$$Similarly, we can reason for any constant function: $f(x) = c$, where $c$ is a constant. So, when we take the derivative of any constant $c$, it is always equal to zero:

$$c' = 0$$Now consider the function $f(x) = x$. We notice that

$$f(x+\triangle x) = x+\triangle x$$and

$$f(x) = x$$

so,

$$f'(x) =\lim\limits_{\triangle x \to 0} \dfrac{f(x+\triangle x)-f(x)}{\triangle x} = \lim\limits_{\triangle x \to 0} \dfrac{x+\triangle x-x}{\triangle x}=\lim\limits_{\triangle x \to 0} \dfrac{\triangle x}{\triangle x} =1$$Therefore, we can conclude that $$x' = 1$$Now consider the function $f(x) = x^2$. Again, we notice that

$$f(x+\triangle x) = (x+\triangle x)^2 = x^2 + 2x\triangle x + (\triangle x)^2\ \text{ and } \ f(x) = x^2,\\ \ \\f'(x) =\lim\limits_{\triangle x \to 0} \dfrac{f(x+\triangle x)-f(x)}{\triangle x} = \lim\limits_{\triangle x \to 0} \dfrac{x^2 + 2x\triangle x + (\triangle x)^2-x^2}{\triangle x} = \\ \ \\ = \lim\limits_{\triangle x \to 0} \dfrac{2x\triangle x + (\triangle x)^2}{\triangle x} = \lim\limits_{\triangle x \to 0} \dfrac{2x + \triangle x}{1} = 2x$$Thus, $$(x^2)' = 2x$$Similarly, we can calculate that $(x^3) '= 3x^2.$ It seems that every time we take a derivative, we take the exponent and bring it down in front to become a coefficient, and then reduce the exponent by one. This is called the power rule:

$$(x^n)' = n\cdot x^{n-1} \ \ n>0$$Indeed, what we calculated above is consistent with this rule: $(x^2)' = 2 \cdot x^{2-1} = 2x$.

Derivatives of polynomials

Let's find a derivative of $f(x) = 5x^4.$ Now we need the constant multiple rule:

$$(c\cdot f)' = c \cdot f'$$When we take a derivative, the exponent that we pull down front must be multiplied by any existing coefficient. So, $$(cx^n)' = c \cdot n \cdot x^{n-1}$$Thus,

$$(5x^4)' = 5 \cdot 4 \cdot x^{4-1} = 20x^3$$

Now consider the following polynomial:

$$f(x) = 3x^3 + 5x^2 + 8x + 1$$Now we need the sum and the difference rule:

$$(f+g)' = f'+g' \ \text{and } (f-g)' = f'-g'$$Then we can write that

$$(3x^3 + 5x^2 + 8x + 1)' = (3x^3)' + (5x^2)' + (8x)' + (1)'= \\ \ \\= 3\cdot3\cdot x^{3-1} + 5 \cdot 2 \cdot x^{2-1} + 8 \cdot 1 + 0 = 9x^2 + 10x + 8$$So, with the power rule, taking the derivative of any polynomial function is actually very simple.

A derivative of a product and a quotient of functions

Suppose we need to find a derivative of a product of two functions: $f\cdot g$. Notice that this is not as simple as just the product of the two derivatives! We need to use the product rule:

$$(f \cdot g)' = f'\cdot g + f \cdot g'$$We can verify that this works by looking at $f(x) = x^2$. We already know that $(x^2)' = 2x$. But let's apply the product rule:

$$(x^2)' = (x\cdot x)' = x'\cdot x + x \cdot x' = 1 \cdot x + x \cdot 1 = 2x$$Let's try using this rule with another function:

$$(3x^4 \cdot 5x^2)' = (3x^4)'\cdot 5x^2+3x^4 \cdot (5x^2)'= \\ \ \\ =12x^3\cdot 5x^2 +3x^4\cdot 10x = 60x^5 + 30x^5 = 90x^5$$Now let's find a derivative of a quotient of two functions: $\dfrac{f}{g}$. We need to use the quotient rule:

$$\left(\dfrac{f}{g}\right)' = \dfrac{f' \cdot g - f \cdot g'}{g^2}$$According to this rule, let's find a derivative of a function $f(x) = \dfrac{x^5+2}{x+4}$:

$$\left(\dfrac{x^5+2}{x+4}\right)' = \dfrac{(x^5+2)'\cdot(x+4)-(x^5+2)\cdot(x+4)'}{(x+4)^2} = \\ \ \\ = \dfrac{5x^4\cdot(x+4)-(x^5+2)\cdot 1}{(x+4)^2} = \dfrac{4x^5+20x^4+2}{(x+4)^2}$$

Conclusion

In this topic, we have learned about some rules, which are helpful when we try to find derivatives of function: the power rule, the sum and difference rules, and the product and quotient rules. In the following topics we will learn how to find derivatives of more complex functions.