**Introduction**

We know that if 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.

**Definition **

The Quotient Rule is a method for determining the derivative (differentiation) of a function in the form of the ratio of two differentiable functions. This means that we can apply the quotient rule when we have to find the derivative of a function of the form $\frac{f ( x )}{g ( x )}$ , such that both f ( x ) and g ( x ) are differentiable, and g ( x ) ≠ 0. Let us now learn the formula of quotient rule.

**Formula of Quotient Rule**

Quotient rule is stated as the ratio of the quantity of the denominator times the derivative of the numerator function minus the numerator times the derivative of the denominator function to the square of the denominator function.

Therefore, if f ( x ) and g ( x ) are differentiable functions and g ( x ) ≠ 0, then

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

Quotient rule is used both in differentiation and integration. Let us learn about them one by one.

**Quotient Rule in Differentiation**

In differentiation, as stated above, the quotient rule is used to find the derivative a function which is of the form f ( x ) and g ( x ) and g ( x ) ≠ 0. The formula in this case is the case as the one defined above for quotient rule, i.e. if f ( x ) and g ( x ) are differentiable functions and g ( x ) ≠ 0, then

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

Let us understand it through an example.

**Example**

Find the derivative of tan x.

**Solution**

We know that tan x = $\frac{sinsin x}{cos x}$

Applying quotient rule to find the derivative of tan x, we get

$\frac{d}{dx} tan tanx = \frac{d}{dx} { \frac{sin sinx}{cos x} }$

⇒ $\frac{d}{dx} tan tanx = \frac{cos x \frac{d}{dx} sinsin x – sinsin x \frac{d}{dx} coscos x}{cos^2x}$

⇒ $\frac{d}{dx} tan tanx = \frac{cos^2x+ sin^2x}{cos^2x}$

⇒ $\frac{d}{dx} tan tanx = \frac{1}{cos^2x} = sec^2x$

**Hence, **$\frac{d}{dx} tantanx = sec^2x$

**Other Fundamental Rules of Differentiation**

Below are some other 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**–**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 ) ].

**Quotient Rule in Integration**

Quotient rule in Integration is known as integration by parts. It is also known as Antiderivative quotient or division rule. Integration by parts is used to integrate the product of two or more functions. The two functions to be integrated f(x) and g(x) are of the form ∫ f ( x ). g ( x ). Thus, it can be called a product rule of integration. Integration by Parts is thus defined as –

The integral of the product of two functions = ( First function ) x ( Integral of second function ) – Integral of { ( Differentiation of first function ) x ( Integral of second function ) }.

This means that if u and v are two functions of x, then

∫ u v d x = u ∫ v d x – { ∫ $\frac{d u}{d x}$ ∫ v d x } d x

It is important to note the following with regards to evaluating an integral by integration by parts –

**Proper choice of first and second function**

Integration with the help of the above rule is called the integration by parts. In the above rule, there are two terms of the right hand side of the function and in both the terms, the integral of the second function is involved. Therefore, in the product of two functions if one of the two functions cannot be integrated directly, for instance in cases of log x, sin ^{– 1} x, cos ^{– 1} x, tan ^{– 1} x etc. we take it as the first function and the remaining function is taken as the second function. If there is no other function, we take unity as the second function. If in the integral both the function can be easily integrated then the first function is chosen in such a manner that the derivate of the function is a simple function and the function thus obtained under the integral sign can be integrated easily than the original function.

**Choosing the first function according to ILATE**

We can also choose the first function as the function which comes first in the word ILATE where,

I – stands for the inverse trigonometric function such as ( sin ^{– 1} x, cos ^{– 1} x , tan ^{ – 1} x etc. )

L – stands for the logarithmic functions

A – stands for the algebraic functions

T – stands for the trigonometric functions

E – stands for the exponential function

Let us understand the above with the help of an example.

**Example**

∫ x sinsin 3 x d x

**Solution**

Here both the functions, i.e. x and sin 3 x can be easily integrated and the derivative of x is one, a less complicated function, therefore, we take x as the first function and sin 3 x as the second function. So, we have,

∫ x sinsin 3 x d x = { x ∫ sinsin 3 x d x – { $\frac{d}{dx}$ ( x ) ∫ sinsin 3 x d x } d x

⇒ ∫ x sinsin 3 x d x = x x ( – $\frac{1}{3}$ ) cos 3 x – ∫ {- $\frac{1}{3}$ cos 3 x } d x

⇒ ∫ x sinsin 3 x d x = – $\frac{1}{3}$ cos 3 x + $\frac{1}{3}$ ∫ coscos 3 x = – $\frac{1}{3}$ cos 3 x + $\frac{1}{9}$ sin 3 x + C

**Hence, ** ∫ x sinsin 3 x d x **= – **$\frac{1}{3}$** cos 3 x + **$\frac{1}{9}$ **sin 3 x + C**

**Solved Examples**

**Example 1** Find the derivative of f ( x ) = $\frac{6x^2}{2-x}$

**Solution **We need to find the derivative of f ( x ) = $\frac{6x^2}{2-x}$. We shall use quotient rule for the same. We have,

$\frac{d}{dx} f ( x ) = \frac{d}{dx} \left\{ \frac{6x^2}{2-x} \right\}$

⇒$\frac{d}{dx} f ( x ) = \frac{(2-x) \frac{d}{dx} 6x^{2} – 6x^{2} \frac{d}{dx} (2-x)}{(2-x)^{2}}$

⇒$\frac{d}{dx} f ( x ) = \frac{12x (2-x) – 6x^2 ( -1 )}{(2-x)^2}$

⇒$\frac{d}{dx} f ( x ) = \frac{24x – 6x^2}{(2-x)^2}$

**Hence, the derivative of the function f ( x ) = **$\frac{6 x^2}{2-x}\: is\: \frac{6 x^2}{2-x}$

**Example 2**** **Find the integral of x^{2}e^{x} by using the integration by parts formula.

**Solution**** **We shall use** **ILATE to solve the given integral. We have,

u = x ^{2} and d v = e ^{x} d x.

Then, d u = 2 x d x, v = ∫ e ^{x} d x = e ^{x}.

Now, using one of the integration by parts formulas, we will have

∫ u d v = u v – ∫ v d u

∫ x ^{2} e ^{x} d x = x ^{2} e ^{x} – ∫ e ^{x} ( 2 x ) d x

= x ^{2} e ^{x} – 2 ∫ x e ^{x} d x

Applying integration by parts formula again to evaluate ∫ x e^{x} dx,

∫ x^{2} e^{x} dx = x^{2} e ^{x} – 2 (x e ^{x} – ∫ e ^{x} dx) = x ^{2} e ^{x} – 2 x e ^{x} + 2 e ^{x} + C

= e ^{x} (x ^{2}– 2 x + 2 )+ C

**Hence, the integral of x**^{2}**e**^{x}** by using the integration by parts formula = e **^{x}** (x **^{2}**– 2 x + 2 )+ C**

**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. - The Quotient Rule – If f ( x ) and g ( x ) are differentiable functions and g ( x ) ≠ 0, then

$\frac{d}{dx} \frac{f ( x )}{g ( x )} = \frac{g ( x ) \frac{d}{dx} [ f ( x ) – f ( x ) \frac{d}{dx} [ g ( x ) ]}{g [ x ]^2}$ - Quotient rule in Integration is known as integration by parts. The integral of the product of two functions = ( First function ) x ( Integral of second function ) – Integral of { ( Differentiation of first function ) x ( Integral of second function ) }. This means that if u and v are two functions of x, then $\int u\: v\: d\: x = u \int v\: d\: x – \left \{ \int \frac{d\: u}{d\: x} \int v\: d\: x \right \} d\: x$