Using the definition of derivatives for solving differentiation, it is quite a long process to solve complex derivatives. In this way, mathematicians solved complex differentiation and made it easy.
Derivative Rules
So, for evaluating these types of complex derivatives we will use the following basic differentiation rules:
Constant Rule Derivative:
$$ \frac{d}{d(x)} (constant)=0 $$
Proof:
We know the derivative gives the slope of a function at any point.
Since the slope of a constant value function is also 0.
So, we can say that the derivative rule of a constant should always be zero.
Example:
Consider a function:
$$ f(x) = 14 $$
Since, constant function has slope=0.
so,
$$ f'(x) = 0 $$
Power Rule of Derivative:
Sometimes there is a differentiable function having a single or multiple term function which have some coefficient in its power. These are called power function, such type of function can be differentiated by power rule of derivative. It is stated as:
$$ \frac{d}{dx}(x^n)=n x^{n-1} $$
Proof:
Function:
$$f(x)= x^n$$ $$ f(a)= a^n $$
Definition of the derivative is:
$$ f'(a) = \lim\limits_{x \to a} \frac{f(x) − f(a)}{x-a} $$ $$ f'(x) = \lim\limits_{x \to a} \frac{x^n- a^n}{Δx} $$
Since,
$$ x^n−a^n = (x−a)(x^{n−1}+x^{n−2} a+⋯+xa^{n−2}+a^{n−1}) $$ $$ f'(a) = \lim\limits_{x \to a} \frac{(x−a)(x^{n−1}+x^{n−2} a+⋯+xa^{n−2}+a^{n−1})}{x-a} $$ $$ f'(a) = \lim\limits_{x \to a}(x^{n−1}+x^{n−2} a+⋯+xa^{n−2}+a^{n−1}) $$ $$ f'(a) =a^{n−1} + a^{n−2}.a +⋯+ a.a^{n−2} + a^{n−1} $$ $$ f'(a) =a^{n−1} + a^{n−1} + .... $$ $$ f'(a) =na^{n−1} $$
In terms of x,
$$ f'(x) =nx^{n−1} $$
Example :
Here we described simple examples which elaborates the power rule of derivative and also make your concept for that rule like iron.
- Consider a function:
- $$ f(x) = x^2 $$ $$ f'(x) = 2x $$
- Consider a function:
- $$f(x) = x^5 $$ $$ f'(x) = 5x^{5-1} $$ $$ f'(x) = 5x^4 $$
Constant Multiple Rule:
If a differentiable function has a constant in it then the derivative of a function will be:
$$ \frac{d}{dx}[cf(x)] = c. \frac{d}{dx}f(x) $$
Here, c = Real number
Proof:
Consider
$$ f(x) = cg(x)$$
Derivative is
$$ f'(x) = \lim\limits_{h \to 0} \frac{f(x+h) − f(x)}{h} $$
Here,
$$ f'(x) = \lim\limits_{h \to 0} \frac{c.g(x+h) −c.g(x)}{h} $$ $$ f'(x) = \lim\limits_{h \to 0} c \left( \frac{g(x+h) − g(x)}{h} \right) $$ $$ f'(x) = c \lim\limits_{h \to 0} \left( \frac{g(x+h) − g(x)}{h} \right) $$ $$ f'(x) = cg'(x) (Proved) $$
Example:
Consider a function:
$$ y= 5x^3 $$ $$ y'= 5 \frac{d}{dx}(x3) $$ $$ y' = 15x^2 $$
Sum and Difference Rule Derivatives:
Consider functions f(x) and g(x), then the sum or difference rule will be:
$$ \frac{d}{dx} (f(x) \pm g(x)) = \frac{d}{dx}f(x) \pm \frac{d}{dx}g(x) $$
Proof:
Consider a function
$$ f(x)= g(x)+h(x) $$ $$ f'(x) = \lim\limits_{a \to 0} \frac{f(x+a) − f(x)}{a} $$ $$ f'(x) = \lim\limits_{a \to 0} \left[ \frac{(g(x+a)+h(x+a)) − (g(x)+h(x))}{a} \right] $$ $$ f'(x) = \lim\limits_{a \to 0} \left[ \frac{g(x+a) − g(x)}{a} + \frac{h(x+a) − h(x)}{a} \right] $$ $$ f'(x) = \lim\limits_{a \to 0} \frac{g(x+a) − g(x)}{a} + \lim\limits_{a \to 0} \frac{h(x+a) − h(x)}{a} $$ $$ f'(x) = g'(x) + h'(x) $$
Similarly, we can prove the difference rule derivatives by opposing signs in the above proof.
Example:
Consider a function:
$$ f(x)= 3x^5-2x^3+9x^2 $$ $$ f'(x)= \frac{d}{dx}(3x^5-2x^3+9x^2) $$ $$ f'(x)= \frac{d}{dx}(3x^5) - \frac{d}{dx}(2x^3) + \frac{d}{dx}(9x^2) $$ $$ f'(x) = 15x^4 -6x^2 + 18x $$
Product Rule of Differentiation:
Product rule tells us hwo to find the derivative of two function which are multiplied to each other. It is the major concept under the topic of derivative rules.
Consider the f(x) and g(x) differentiable functions, then the product rule of Differentiation will be stated as:
$$ \frac{d}{dx}[f(x) \cdot g(x)] = f(x) \frac{d}{dx}[g(x)] + g(x) \frac{d}{dx}[f(x)] $$
or
$$ \frac{d}{dx}[f(x) \cdot g(x)] = f(x)g'(x) + g(x)f'(x) $$
Proof of Product Rule:
Consider a function:
$$ u(x)= f(x) \cdot g(x) $$
Solution:
$$ u'(x) = \lim\limits_{h \to 0} \frac{u(x+h) − u(x)}{h} $$ $$ \implies u'(x) = \lim\limits_{h \to 0} \frac{f(x+h)g(x+h)−f(x)g(x)}{h} $$
We have to add and subtract in f(x+h)g(x) numerator
$$ u'(x) = \lim\limits_{h \to 0} \left[ \frac{f(x+h)g(x+h)-f(x+h)g(x)+f(x+h)g(x)−f(x)g(x)}{h} \right] $$ $$ u'(x) = \lim\limits_{h \to 0} \left[ \{ f(x+h) \left(\frac{g(x+h)−g(x)}{h} \right) \} + \{ g(x) \left(\frac{f(x+h)−f(x)}{h} \right) \} \right] $$ $$ \implies u'(x) =f(x)g'(x) + g(x)f'(x) $$
Example:
here is the example for proper understanding of product rule of differentiation.
Consider a function:
$$ f(x) = (2x^2)(3x+2) $$
Solution:
$$ f'(x) = (2x^2) \frac{d}{dx}(3x+2) + (3x+2)\frac{d}{dx}(2x2) $$ $$ f'(x) = (2x^2)(3)+(3x+2)(4x) $$ $$ f'(x) = 6x^2 + 12x^2 + 8x $$ $$ f'(x) = 18x^2 + 8x $$
Quotient Rule:
Product and quotient rule both are the most important rules of differentiation which are furthur used in many important theorems of derivatives and integrations.
Product rule of differentiation is the base rule used in the integration by parts. Similarly quotient rule also used in many other theorems.
Consider the f(x) and g(x) differentiable functions, then the Quotient rule formula will be stated as:
$$ \frac{d}{dx} \left[ \frac{f(x)}{g(x)} \right] = \frac{g(x)\frac{d}{dx}[f(x)]−f(x) \frac{d}{dx}[g(x)]}{[g(x)]^2} $$
Proof of Quotient Rule:
Consider a function:
$$ u(x)= \frac{f(x)}{g(x)} $$
Solution:
$$ u'(x) = \lim\limits_{h \to 0} \frac{u(x+h) − u(x)}{h} $$ $$ \implies u'(x) = \lim\limits_{h \to 0} \frac{ \frac{f(x+h)}{g(x+h)}− \frac{f(x)}{g(x)}}{h} $$ $$ u'(x) = \lim\limits_{h \to 0} \frac{f(x+h)g(x)-g(x+h)f(x)}{h \cdot g(x)g(x+h)} $$ $$ u'(x) = \lim\limits_{h \to 0} \left( \frac{1}{g(x)g(x+h)} \right) \lim\limits_{h \to 0} \left( \frac{f(x+h)g(x)-g(x+h)f(x)}{h} \right) $$ $$ u'(x) = \left[ \frac{1}{[g(x)]^2} \right] \lim\limits_{h \to 0} \left[ \frac{f(x+h)g(x)-g(x+h)f(x)}{h} \right] $$
We have to add and subtract in f(x)g(x) in the numerator of limit
$$ u'(x) = \left[ \frac{1}{[g(x)]^2} \right] \lim\limits_{h \to 0} \left( \frac{f(x+h)g(x)-f(x)g(x)+f(x)g(x)-g(x+h)f(x)}{h} \right) $$ $$ u'(x) = \left[ \frac{1}{[g(x)]^2} \right] \lim\limits_{h \to 0} \left[ g(x) \frac{f(x+h)-f(x)}{h}- f(x) \frac{g(x+h)-g(x)}{h} \right] $$ $$ u'(x) = \left[ \frac{1}{[g(x)]^2} \right] \left[ g(x) \lim\limits_{h \to 0} \frac{f(x+h)-f(x)}{h} - f(x) \lim\limits_{h \to 0} \frac{g(x+h)-g(x)}{h} \right] $$ $$ \implies u'(x) =\left[ \frac{1}{[g(x)]^2} \right] g(x)f'(x)-f(x)g'(x) $$ $$ u'(x) = \frac{g(x)f'(x)-f(x)g'(x)}{[g(x)]^2} $$
Example:
Quotient rule derivatives tells us how to find the derivative of a fraction. Let's see amazing examples of quotient rule.
Consider a function:
$$ f(x)= \frac{3x+2}{2x^2} $$
Solution:
$$ f'(x)= \frac{2x^2 \frac{d}{dx}(3x+2) - (3x+2)\frac{d}{dx}(2x^2)}{2x^2} $$ $$ f'(x)= \frac{2x^2(3) - (3x+2)(4x)}{2x^2} $$ $$ f'(x)= \frac{6x^2 - (12x^2+8x)}{2x^2} $$ $$ f'(x)= \frac{-6x^2 - 8x} {2x^2} $$ $$ f'(x)=(-2x)( \frac{3x + 4}{2x^2}) $$ $$ f'(x) = -\frac{3x+4}{x} $$
So, these are some basic and important rules for finding the derivative of the function.