Derivative Rules

Derivatives are important concepts of Mathematics. Derivatives are basic to the different solutions to the problems of calculus and differential equations. Generally, scientists observe a dynamic system to get the rate of change of some variable of interest, including this information into some differential equation and use integration methods to obtain functions that can be used to estimate the behaviour of the original system in different conditions. Let us now discuss what is derivative in Mathematics?

Derivatives in Mathematics is the rate of change of a function in terms of a variable. The rate of change of a function in derivatives can be estimated by calculating the ratio of the change of the function $\Delta b$ to the change of the independent variable $a$. This ratio in the derivative is considered in the limit as $\Delta a \to 0$.

As you have learned what is derivative in Mathematics, let us now discuss how to define derivatives and derivative rules that can be used to calculate many derivatives.

Define Derivatives

Let $fx$ be a function whose domain includes an open interval at some point $x_0$. Then the function $fa$ is considered to be differentiable at $x_0$, and the derivative of $fx$ at $x_0$ is expressed as:

$f’ x_0 = \underset{\Delta \to 0}{lim}{\dfrac{\Delta y}{\Delta x}}$

$\Rightarrow \underset{\Delta \to 0}{lim}{\dfrac{f(x_0 + \Delta x - f(x_0)}{\Delta x}}$

The derivative of a function $y$ in Lagarangee’s form is expressed as:

$y = f(x) \text{ as } f '(x) \text{ or } y' (x)$

The derivative of a function y in Leibniz’s form is expressed as:

$y = f (x) \text{ as } \dfrac{df}{dx} \text{ or } \dfrac{dy}{dx}$

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Derivative Rules

Below are some of the derivative rules that can be used to calculate differentiation questions.

The Constant Rule

Let $y$ be an arbitrary real number. The constant rule is defined as:

$\dfrac{d(y)}{dx} = 0$

The Constant Function Rule

Let $y$ be an arbitrary real number, and $g(x)$ be an arbitrary differentiable function. The constant function rule states that

$\dfrac{d(y \cdot g(x))}{dx} = y \cdot g’(x)$

The Power Rule

Let $a$ and $b$ be a real number, with $a \neq 0$ and $a$ and $b$. The Power rule states that

\[\frac{d}{dx} x^{n} = n x^{n-1}\]

The Product Rule

Let $a(x)$ and $b(x)$ be an arbitrary differentiable function. The product rule states that

$\dfrac{d(a(x) \cdot b(x))}{dx} = a’(x) \cdot b(x) + a(x) + b’(x)$

The Chain Rule

The derivative of the function $h(x)= a(b(x))$  in terms of chain rule is expressed as:

$h'(x)= a'(b(x)) \cdot  b'(x)$.

The product rule in Leibniz's notation is represented as

$\dfrac{dh(x)}{dx} = \dfrac{da(b(x))}{db(x)} \cdot \dfrac{db(x)}{dx}$

The $x$ Rule

Let us consider $x$ as an arbitrary variable, then $X$ rule states that

$\dfrac{d(x)}{dx} = 1$

The Sum and Difference Rule

Let $a(x)$ and $b(x)$ be an arbitrary differentiable function. 

Recall that for an arbitrary function $k(x)$,

$\dfrac{d(k(x)}{dx} = k’(x) = \underset{\Delta h \to 0}{lim} \dfrac{k(x+h)-k(x)}{h}$

The sum rule states that:

$\dfrac{d(a(x)+b(x))}{dx} = a’(x)+b’(x)$

The difference rule states that:

$\dfrac{d(a(x)-b(x))}{dx} = a’(x) - b’(x)$

The Quotient Rule

Let $a(x)$ and $b(x)$ be an arbitrary differentiable function with $a(x) \neq 0$ and $b(x) \neq 0$. The quotient rule states that:

$\dfrac{d\left({\dfrac{a(x)}{b(x)}}\right)}{dx} = \dfrac{a’(x) \cdot b(x) - a(x) \cdot b’(x)}{(b(x))^2}$

Fun Facts

• Gottfried Wilhelm Leibniz introduced the symbols $dx, dy$ and $\dfrac{dy}{dx}$ in 1675. The symbols are commonly used when the equation $y = f(x)$ is examined as a functional relation between dependent and independent variables.

• The first derivative is represented by $\dfrac{dy}{dx}, \dfrac{df}{dx}$ or $\dfrac{d}{dx} f$, and was once considered as an infinitesimal quotient.

Solved Examples

1. Evaluate $\dfrac{d}{dx}( 2x + 1)^2$ using the chain rule.

Solution:

Let $g(x)= (2x + 1)$, and $f(x)= x^2$

Then, $f(g(x))=  (2x + 1)^2$

As we know $f '(x)= 2x$, and $g'(x)= 2$.

Accordingly, $\dfrac{d}{dx}( 2x + 1)^2 = f'(g(x)) \cdot g'(x)$

$= f'( 2x + 1) \cdot 2$

$= 2 ( 2x + 1) \cdot 2$

$= 8x + 4$

2. Differentiate the function $f(x) = x^{10}$ using power rule.

Solution:

$f'(x)= 10x^{10-1}$ 

$= 10x^9$

3. Find the derivative of the following function:

$y = \dfrac{1-y}{y^2 + 2}$

Solution:

We have

$y’ = \dfrac{( 1 -k)' ( k^2 + 2) - ( 1 - k) (k^2 + 2)'}{(k^2 + 2)^2}$

$y’ =  \dfrac{(-1) \cdot (k^2 + 2)^2 - ( 1 - k)(k^2 + 2)^{2’}}{(k^2 + 2)^2}$

$y’= k^2 - 2k - 2(k^2 + 2)^2$