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How to Find the Derivative

August 30, 2021 Math Olympiads Topics, Math Topics
In order to find the derivative $f'(x)$ of a particular function $f(x)$ we need to know the following:
    • The derivatives of the basic functions (like $x^2$, $e^x$, $\cos{x}$, etc.)
    • The rules of differentiation for functions that have operations (addition, subtraction, multiplication, etc.)

Derivatives of the Basic Functions

The derivatives of the basic functions are given in the following table (here $c \in \mathbb{R}$, $a>0$, $a \neq 1$).

$f(x)$  $f'(x)$
$c$ 
$0$
$x$
 $1$
$x^2$
 $2x$
$x^3$
$3x^2$
$\frac{1}{x}$
$-\frac{1}{x^2}$
$\sqrt{x}$ 
$\frac{1}{2\sqrt{x}}$
$\sqrt[3]{x}$
$\frac{1}{3\sqrt[3]{x^2}}$
 $x^n$
$nx^{n-1}$

 


 

$\sin{x}$
$\cos{x}$
$\cos{x} $
$-\sin{x}$
$\tan{x}$
$\sec^2{x}$
$\cot{x}$
$-\csc^2{x}$
$\sec{x}$
$\sec{x}\tan{x}$
$\csc{x}$ 
$-\csc{x}\cot{x}$

 


 

$e^{x}$ 
$e^{x}$
$a^{x}$
$a^{x} \cdot \ln{a}$
$\ln{x}$
 $\frac{1}{x}$
$\log_{a}{x}$
$\frac{1}{x \cdot \ln{a}}$

 


 

$\sin^{-1}{x}$
$\frac{1}{\sqrt{1-x^2}}$
$\cos^{-1}{x}$
$-\frac{1}{\sqrt{1-x^2}}$
$\tan^{-1}{x}$ 
$\frac{1}{1+x^2}$
$\cot^{-1}{x}$
$-\frac{1}{1+x^2}$
$\sec^{-1}{x}$
$\frac{1}{|x|\sqrt{1-x^2}}$
$\csc^{-1}{x}$
$-\frac{1}{|x|\sqrt{1-x^2}}$
 
 



Rules of Differentiation

For the functions that involve operations on other functions, there are six rules of differentiation: Coefficient Rule, Addition Rule, Subtraction Rule, Product Rule, Quotient Rule, and Chain Rule.

The Coefficient Rule states that for a constant $k \in \mathbb{R}$ and the function $f(x)$

$ \left( k \cdot f(x) \right)’= k \cdot f'(x) $

Example

$ \left( 3 \cdot \sin{x} \right)’ = 3 \cdot \left( \sin{x} \right)’ = \boxed{3\cos{x}} $

The Addition Rule states that for the functions $f(x)$ and $g(x)$

$ \left( f(x)+g(x) \right)’= f'(x) + g'(x) $

Example

$ \left( x^2 + e^{x} \right)’ = \left( x^2 \right)’ + \left( e^{x} \right)’ = \boxed{ 2x+e^x} $

The Subtraction Rule states that for the functions $f(x)$ and $g(x)$

$ \left( f(x)-g(x) \right)’= f'(x) – g'(x) $

Example

$ \left( x^3 – \ln{x} \right)’ = \left( x^3 \right)’ – \left( \ln{x} \right)’ = \boxed{ 3x^2- \frac{1}{x}} $

The Product Rule states that for the functions $f(x)$ and $g(x)$

$ \left( f(x) \cdot g(x) \right)’= f'(x) \cdot g(x) + g'(x) \cdot f(x) $

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Example

$ \left( x \cdot \tan{x} \right)’ = \left( x \right)’ \cdot \left( \tan{x} \right) + \left( \tan{x} \right)’ \cdot \left( x \right) = \left( 1 \right) \cdot \left( \tan{x} \right) + \left( \sec^2{x} \right) \cdot \left( x \right) = \boxed{ \tan{x} + x \sec^2{x}} $

The Quotient Rule states that for the functions $f(x)$ and $g(x)$

$ \left( \frac{f(x)}{g(x)} \right)’= \frac{f'(x) \cdot g(x) – g'(x) \cdot f(x)}{g^2(x)} $

Example

$ \left( \frac{2x-1}{x} \right)’ = \frac{(2x-1)’ \cdot (x) – (x)’ \cdot (2x-1) }{x^2} = \frac{(2) \cdot (x) – (1)\cdot (2x-1) }{x^2} = \boxed{ \frac{1}{x^2} } $
 

The Chain Rule states that for the functions $f(x)$ and $g(x)$

$ \left( f(g(x)) \right)’= f'(g(x)) \cdot g'(x) $

Example

$ \sin \left( x^2+3x \right)’= \cos \left(  x^2+3x \right) \cdot \left(  x^2+3x \right)’ = \boxed{ \left( 2x+3 \right) \cos \left(  x^2+3x \right) } $

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