Derivatives in Maths | Definition, Examples, Rules, Derivative of a Function

Definition

Let f ( x ) be a real valued function defined on an open interval ( a, b ) and let c ∈ ( a, b ). Then, f ( x ) is said to be differentiable or derivative at x = c if and only if

$\frac{f (x) -f ( c )}{x-c}$ exists finitely.

This limit is called the derivative or differentiation of f ( x ) at x = c and is denoted by f ‘ ( c ) or D f ( c ) or { $\frac{d}{dx}$ f ( x ) } _{x = c}

This means that

f ‘ ( c ) =$\frac{f (x) -f ( c )}{x-c}$ provided that the limit exists

Let us understand it through an example

Example

Find the derivative of f ( x ) = k at x = 0 and x = 5

Solution

We are required to find the derivative of f ( x ) = k at x = 0 and x = 5

Now,

f ‘ ( 0 ) = $\frac{f (0 + h) -f ( 0 )}{h} = \frac{k-k}{h} = \frac{0}{h} = 0$

Similarly,

f ‘ ( 5 ) = $\frac{f (5 + h) -f ( 0 )}{h} = \frac{k-k}{h} = \frac{0}{h} = 0$

Derivative of a Function

Let f ( x ) be a function differentiable at every point in its domain. Then corresponding to every point c in the domain, we obtain a unique real number equal to the derivative f ‘ ( c ) of f ( x ) at x = c.

This implies that there is a one – one correspondence between points in the domain of the function and the derivative at these points. This correspondence induces a function such that the image of any point x in the domain is the value of the derivative of f at x, i.e. f ‘ ( x ) of $\frac{d}{dx}$ f ( x ) . This function is called the derivate of differentiation of f ( x ) and is given by

f ‘ ( x ) = $\frac{f (x + h) -f ( x )}{h}$ or

$\frac{d}{dx} f ( x ) = \frac{f (x + h) -f ( x )}{h}$

The above process of finding the derivative of a function is known as the differentiation or derivative from the first principles. Let us learn more about it.

Differentiation from the First Principles

We have learned that the derivative of a function f ( x ) is given by

$\frac{d}{dx} f ( x ) = \frac{f (x + h) -f ( x )}{h}$

Let us now look at the derivatives of some important functions –

The Power Rule – If f ( x ) = x ⁿ, where n ∈ R, the differentiation of x ⁿ with respect to x is n x ^{n – 1} therefore,

$\frac{d}{dx}$ ( x ⁿ ) = n x ^{n – 1}

Let us understand it through an example

Example

Find the value of $\frac{d}{dx}$( x ⁵ )

Solution

We have been asked to find the value of $\frac{d}{dx}$( x ⁵ )

We have,

$\frac{d}{dx}$( x ⁵ ) = 5 x ^{5 – 1} = 5 x ⁴

Hence, $\frac{d}{dx}$( x ⁵ ) = 5 x ⁴

If f ( x ) = x ⁿ, where n R, the differentiation of x ⁿ with respect to x is n x ^{n – 1} therefore,
If f ( x ) = e ^x, then the differentiation of e ^x with respect to x is e ^x.
If f ( x ) = x , where x > 0 then the differentiation of x with respect to x is $\frac{1}{x}$.
If f ( x ) = sin x, then the differentiation of sin x with respect to x is cos x.
If f ( x ) = cos x, then the differentiation of cos x with respect to x is – sin x.
If f ( x ) = tan x, then the differentiation of tan x with respect to x is sec ² x.
If f ( x ) = cot x, then the differentiation of cot x with respect to x is – cosec ² x.
If f ( x ) = sec x, then the differentiation of sec x with respect to x is sec x tan x.
If f ( x ) = cosec x, then the differentiation of cosec x with respect to x is – cosec x cot x.

Fundamental Rules of Differentiation

Now, that we have learnt the basic of derivatives, let us move to understanding some fundamental rules of differentiation.

Below are some fundamental rules of differentiation –

The Constant Rule – Differentiation of constant function is zero, i.e. $\frac{d}{dx}$ ( c ) = 0. For example, if we have a function f (x) = 5, we can see that it is a horizontal line with a slope of zero, and thus its derivative is also zero.
The Constant Multiple rule – Let f ( x ) be a differentiable function and let c be a constant. The, c f ( x ) is also differentiable such that $\frac{d}{dx}$ [ c f ( x ) ] = c $\frac{d}{dx}$( f ( x ) ).
The Sum Rule – If f ( x ) and g ( x ) are differentiable functions, then f ( x ) + g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) + g ( x ) } = $\frac{d}{dx}$[ f ( x ) ] + $\frac{d}{dx}$[ g ( x ) ]
The Difference Rule – If f ( x ) and g ( x ) are differentiable functions, then f ( x ) – g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) – g ( x ) } = $\frac{d}{dx}$[ f ( x ) ] – $\frac{d}{dx}$[ g ( x ) ]
The Product Rule – If f ( x ) and g ( x ) are differentiable functions, then f ( x ) g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) g ( x ) } = f ( x ) $\frac{d}{dx}$[ g ( x ) ] + g ( x ) $\frac{d}{dx}$[ f ( x ) ].
The Quotient Rule – If f ( x ) and g ( x ) are differentiable functions and g ( x ) ≠ 0, then

$\frac{d}{dx} \frac{f ( x )}{g ( x )} = \frac{g ( x ) \frac{d}{dx}[ f x – f ( x ) \frac{d}{dx}[ g ( x ) ]}{g\left [(x)\right]^2}$

Examples

Example 1 Find the derivative of the function f ( x ) = 6 x ² – 4 x.

Solution We have been given the function. We can clearly see that the function is of the form f ( x ) = u ( x ) – v ( x )

Now, we shall recall the difference rule that states that If f ( x ) and g ( x ) are differentiable functions, then f ( x ) – g ( x ) is also differentiable such that

$\frac{d}{dx} [ f ( x ) – g ( x ) = \frac{d}{dx}[ f ( x ) ] – \frac{d}{dx} [ g ( x ) ]$

Using this rule in the given function, we will get

$\frac{d}{dx} [ 6 x^2 – 4 x ] = \frac{d}{dx}( 6 x^2) – \frac{d}{dx}( 4 x )$

$\frac{d}{dx}$[ 6 x ² – 4 x ] = 6 ( 2 x ) – 4

$\frac{d}{dx}$[ 6 x ² – 4 x ] = = 12 x – 4

Hence, the derivative of the function f ( x ) = 6 x ² – 4 x will be 12 x – 4

Example 2 Differentiate f ( x ) = ( x + 2 ) ³/ √x

Solution We have been given the function f ( x ) = ( x + 2 ) ³/ √x

⇒ f ( x ) = ( x + 2 ) ( x ²+ 4 x + 4 ) / √x

⇒ f ( x ) = [ x ³+ 6 x ²+ 1 2 x + 8 ] / x ^½

⇒ f ( x ) = x ^-1/2( x ³+ 6 x ²+ 1 2 x + 8 )

⇒ f ( x ) = x ^{5 / 2}+ 6 x ^{3 / 2}+ 12 x ^{1 / 2}+8 x ^{– ½}

Now, we can see that the given function is of the form where we can apply the sum rule.

Now, we shall recall that if f ( x ) and g ( x ) are differentiable functions, then f ( x ) + g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) + g ( x ) } = $\frac{d}{dx}[ f ( x ) ] + \frac{d}{dx}[ g ( x ) ]$

Therefore, differentiating the given equation using sum rule, we get;

$\frac{d}{dx}$(x ^{5 / 2}+ 6 x ^{3 / 2}+ 12 x ^{1 / 2}+8 x ^{– ½} )

= 5 / 2 x ^{3 / 2} + 6 ( 3 / 2 x ^{1 / 2}) + 1 2 ( 1 /2 x ^{– 1 / 2}) + 8 ( − 1 / 2 x ^{– 3 / 2})

= 5 / 2 x ^{3 / 2}+ 9x ^{1 / 2} + 6 x ^{– ½} − 4 x^{– 3 / 2}

Hence, $\frac{d}{dx}$( x + 2 ) ³/ √x = 5 / 2 x ^{3 / 2}+ 9x ^{1 / 2} + 6 x ^{– ½} − 4 x^{– 3 / 2}

Example 3 Find the derivative of x ² ( x + 3 )

Solution We have been given the function f ( x ) = x ² ( x + 3 )

Now, we can see that the given function is of the form where we can apply the product rule.

Now, we shall recall that if f ( x ) and g ( x ) are differentiable functions, then f ( x ) g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) g ( x ) } = f ( x ) $\frac{d}{dx}$ [ g ( x ) ] + g ( x ) $\frac{d}{dx}$ [ f ( x ) ].

Here,

f ( x ) = x ²and g ( x ) = x+3

Therefore, differentiating the given equation using product rule, we get;

$\frac{d}{dx}$[ x ²( x + 3 ) ] = x ²$\frac{d}{dx}$( x + 3 ) + ( x + 3 ) $\frac{d}{dx}$ x ²

$\frac{d}{dx}$[ x ²( x + 3 ) ] = x ²( 1 ) + 2 x ( x + 3 )

$\frac{d}{dx}$[ x ²( x + 3 ) ] = 2 x ²+ 6 x + x ²

$\frac{d}{dx}$[ x ²( x + 3 ) ] = 3 x ²+ 6 x

$\frac{d}{dx}$[ x ²( x + 3 ) ] = 3 x ( x + 2 )

Hence, $\frac{d}{dx}$[ x ²( x + 3 ) ] = 3 x ( x + 2 )

Key Facts and Summary

Let f ( x ) be a real valued function defined on an open interval ( a, b ) and let c ∈ ( a, b ). Then, f ( x ) is said to be differentiable or derivative at x = c if and only if
$\frac{f( x ) – f( c )}{x-c}$ exists finitely.
If f ( x ) = e ^x, then the differentiation of e ^x with respect to x is e ^x.
If f ( x ) = x , where x > 0 then the differentiation of x with respect to x is $\frac{1}{x}$.
If f ( x ) = sin x, then the differentiation of sin x with respect to x is cos x.
If f ( x ) = cos x, then the differentiation of cos x with respect to x is – sin x.
If f ( x ) = tan x, then the differentiation of tan x with respect to x is sec ² x.
If f ( x ) = cot x, then the differentiation of cot x with respect to x is – cosec ² x.
If f ( x ) = sec x, then the differentiation of sec x with respect to x is sec x tan x.
If f ( x ) = cosec x, then the differentiation of cosec x with respect to x is – cosec x cot.
The Constant Rule – Differentiation of constant function is zero, i.e. $\frac{d}{dx}$( c ) = 0. For example, if we have a function f (x) = 5, we can see that it is a horizontal line with a slope of zero, and thus its derivative is also zero.
The Constant Multiple rule – Let f ( x ) be a differentiable function and let c be a constant. The, c f ( x ) is also differentiable such that $\frac{d}{dx}$[ c f ( x ) ] = c $\frac{d}{dx}$( f ( x ) ).
The Sum Rule – If f ( x ) and g ( x ) are differentiable functions, then f ( x ) + g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) + g ( x ) } = $\frac{d}{dx}$[ f ( x ) ] + $\frac{d}{dx}$[ g ( x ) ]
The Difference Rule – If f ( x ) and g ( x ) are differentiable functions, then f ( x ) – g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) – g ( x ) } = $\frac{d}{dx}$[ f ( x ) ]- $\frac{d}{dx}$[ g ( x ) ]
The Product Rule – If f ( x ) and g ( x ) are differentiable functions, then f ( x ) g ( x ) is also differentiable such that $\frac{d}{dx}$[ f ( x ) g ( x ) } = f ( x ) $\frac{d}{dx}$[ g ( x ) ] + g ( x ) $\frac{d}{dx}$[ f ( x ) ].
The Quotient Rule – If f ( x ) and g ( x ) are differentiable functions and g ( x ) ≠ 0, then

$\frac{d}{dx} \frac{f ( x )}{g ( x )} = \frac{g ( x ) \frac{d}{dx}[ f x – f ( x ) \frac{d}{dx}[ g ( x ) ]}{g\left [(x)\right]^2}$

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