# Derivative Properties and Formulas for Class 11

In this post we will discuss important properties and formulas of differentiation which comes under Class 11 Mathematics NCERT/CBSE Syllabus. Along with the formulas we have also made an effort to derive some of the formulas so that you have an idea from where these formulas have come from.
I urge you to remember each of the formulas as without the you won’t be able to solve the worksheet problems of this chapter.

## Algebra of Derivative of Functions

Let f(x) and g(x) are the two functions with common domain. Some of the properties of algebra of derivatives are given below:

(a) Sum of Derivatives:
The derivative of sum of two function is equal to the the sum of individual derivative of the function

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

(b) Subtraction Property of Derivative
The derivative of difference of two function is equal to the difference of individual derivative of the function

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

(c) Product of Derivative
Use the following product rule to perform the product of any two derivative

\frac{d}{dx}[ f( x) \ .\ g( x)] \\ \\ ⟹ \frac{d}{dx} \ [ f( x)] .g( x) \ +f( x) \ .\frac{d}{dx} \ [ g( x)] \\\ \\

(d) Quotient Rule of Derivative
Derivative of quotient of two functions is given by the following quotient rule (whenever the denominator is non–zero)

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

### Important Derivative formula for Class 11 Calculus

Before getting into the formulas of derivative, remember the following note:
\frac{d}{dx} f( x) \ can\ also\ be\ written\ as\ f'( x)

1. Basic Differentiation of Function with Power
if\ f( x) \ =\ x^{n} \\ \\ then\ f'( x) \ =n.x^{n-1} \\\ \\

2. Derivative of Trigonometry Functions

(a) f( x) \ =\ sin( x)\\ \\ then\ f'( x) \ =cos( x) \\

(b) (cos x)’ = – sin x

(c) (tan x)’ = \frac{1}{cos^{2} x}

(d) (cot x)’ = \frac{1}{-\ sin^{2} x}

(e) (sec x)’ = sec x . tan x

(f) (cosec x)’ = cosec x . cot x

3. Derivative of Constant is Zero

\frac{d}{dx} \ constant\ =\ 0\

4. Derivative of Function with constant value

\frac{d}{dx} \ c.f( x) \ =c\ .\frac{d}{dx} f( x)

5. Differentiation of Log and Exponential Function

(a) \frac{d} {dx} e^{x} =e^{x} \\\ \\

(b) \frac{d}{dx} \ log( x) \ =\frac{1}{x} \\\ \\

(c) \frac{d}{dx} \ log_{a} x\ =\ \frac{1}{x} \ .\ \frac{1}{log\ a} \\\ \\

(d) \frac{d}{dx} \ a^{x} \ =a^{x} .\ log\ a \\\ \\

(e) \frac{d}{dx} \ x^{x} \ =x^{x} \ ( 1+log\ x)

### Formula Derivation

(01) Prove the following differentiation formula
(sin x)’ = cos x

f( x) \ =\ sin( x)\\\ \\ The\ basic\ derivative\ formula\ is:\\\ \\ \frac{d}{dx\ } f( x) \ =\ \lim_ {\Delta x\rightarrow 0}\frac{f( x+\Delta x) -f( x)}{\Delta x}\\\ \\ \\\ \\ \ Putting\ the\ value\ of\ f( x)\\\ \\ \frac{d}{dx} \ sin( x) \ =\lim_{\Delta x\rightarrow 0}\frac{sin( x+\Delta x) -sinx}{\Delta x}\\\ \\ \\\ \\ We\ know\ that \\ \\ Sin\ A\ -\ Sin\ B\ =\ 2\ Cos\frac{A+B}{2} .\ Sin\frac{A-B}{2}\\\ \\ \\\ \\ \frac{d}{dx} \ sin( x) \ =\lim_ {\Delta x\rightarrow 0}\frac{2\ Cos\ \frac{2x+\Delta x}{2} \ .\ Sin\frac{\Delta x}{2}}{\Delta x}\\\ \\ \\\ \\ \ Using\ Limit\ Properties\\ \\ The\ equation\ can\ be\ written\ as\ \\\ \\ \ \frac{d}{dx} \ sin( x) \\ \\ = \lim_ {\Delta x\rightarrow 0} Cos\frac{2x+\Delta x}{2} \ \times \lim_{\Delta x\rightarrow 0}\frac{Sin\frac{\Delta x}{2}}{\frac{\Delta x}{2}}\\\ \\ \\\ \\ We\ know\ that\ \ \Longrightarrow \lim_{x\rightarrow 0} =\ \frac{sin\ x}{x} \ =\ 1\\\ \\ \\\ \\ Using\ the\ Limits\ we\ get\\\ \\ \frac{d}{dx} \ sin( x) \ =cos\ x\ \times \ 1\\\ \\ \frac{d}{dx} \ sin( x) \ =cos\ x\\\ \\

Hence the derivative of sin(x) is cos(x)

(02) Prove the following differentiation formula
\frac{d}{dx} \ tanx\ =sec^{2} x

f( x) \ =\ tan( x)\\\ \\ The\ basic\ derivative\ formula\ is:\\\ \\ \frac{d}{dx\ } f( x) \ =\ \lim_{\Delta x\rightarrow 0}\frac{f( x+\Delta x) -f( x)}{\Delta x}\\\ \\ \ \ Putting\ the\ value\ of\ f( x)\\ \\ ⟹ \lim_{\Delta x\rightarrow 0}\frac{tan( x+\Delta x) -tanx}{\Delta x} \ \ -eq( 1) \\\ \\ We\ know\ that\\ \\ tan\ x\ =\ \frac{sin\ x}{cos\ x}\\\ \\ \\\ \\ Putting\ this\ value\ in\ equation\ ( 1)\\ \\ ⟹ \lim_{\Delta x\rightarrow 0}\frac{1}{\Delta x} \times \left(\frac{sin( x+\Delta x)}{cos( x+\Delta x)} -\frac{sin\ x}{cos\ x}\right)\\\ \\ ⟹ \lim_{\Delta x\rightarrow 0}\frac{1}{\Delta x} \times \left(\frac{sin( x+\Delta x) .cos\ x\ -\ cos( x+\Delta x) .sin\ x}{cos\ x\ \ cos( x+\Delta x)}\right)\\\ \\ We\ know\ that\\ \\ sin\ ( A-B) \ =\ sinA\ cosB\ -\ cos\ A.\ sin\ B\\\ \\ \\\ \\ Using\ the\ formulas\ in\ equation\\\ \\ ⟹ \lim_{\Delta x\rightarrow 0}\left(\frac{sin\Delta x}{\Delta x.cos\ x\ \ cos( x+\Delta x)}\right)\\\ \\ ⟹ \lim_ {\Delta x\rightarrow 0}\frac{sin\Delta x}{\Delta x} \ \times \ \lim _{\Delta x\rightarrow 0}\frac{1}{cosx.\ cos( x+\Delta x)}\\\ \\ Using\ Limits\\ \\ \frac{d}{dx} \ tanx\ =\ 1\ \times \ \frac{1}{cos\ x.\ cos\ x}\\\ \\ \frac{d}{dx} \ tanx\ =sec^{2} x \\\ \\

Hence Proved

### Questions on Differentiation

(01) Find the derivative of following expression
\frac{x+1}{x}

Solution
In this question we will use quotient rule of derivative
If u and v are two functions then according to quotient rule:
\frac{d}{dx}\left(\frac{u}{v}\right) =\frac{u'v-uv'}{v^{2}} \\ \\

Here
u = x + 1
u’= 1

v = x
v’= 1

Putting all these values in the above formula we get

\frac{d}{dx}\left(\frac{x+1}{x}\right) =\frac{1.x-( x+1) .1}{x^{2}}\\\ \\ \frac{d}{dx}\left(\frac{x+1}{x}\right) =\frac{x-x-1}{x^{2}}\\\ \\ \frac{d}{dx}\left(\frac{x+1}{x}\right) =\frac{-1}{\ x^{2}}\\\ \\

Hence \frac{-1}{\ x^{2}}\ is the right answer

(02) Find the derivative of following expression at x = 1
1\ +\ x\ +\ x^{2} +\ x^{3}\ + \ x^{4} +\ \ .\ .\ .\ .\ .\ .x^{50}

Solution
For the above expression we will use the following formula:
if\ f( x) \ =\ x^{n} \\ \\ then\ f'( x) \ =n.x^{n-1} \\\ \\

Using the above formula, let us find the derivative of individual components
\frac{d}{dx} \ 1\ =\ 0 \\\ \\ \frac{d}{dx} \ x\ =\ 1\\\ \\ \frac{d}{dx} \ x^{2} \ =\ 2x\\\ \\ \frac{d}{dx} \ x^{3} \ =\ 3x^{2}\\\ \\ \frac{d}{dx} \ x^{49} \ =\ 49x^{48}\\\ \\ \frac{d}{dx} \ x^{50} \ =\ 50x^{49}\\\ \\

Combining these derivative values as per the given question we get:
0+1+2x+3x^{2} +4x^{3} +\ .\ .\ .\ .\ .49\ x^{48\ } +50\ x^{49} \\\ \\ Putting\ x=1\ in\ the\ above\ equation\\ \\ we\ get\\\ \\ 1+2+3+\ .\ .\ .\ .+\ 49+50\\\ \\ formula\ for\ sum\ of\ series\ of\ number\\\ \\ \Longrightarrow \ \frac{n( n+1)}{2}\\\ \\ Here\ n=50\\\ \\ Putting\ the\ value\\ \\ \Longrightarrow \frac{50\times 51}{2}\\\ \\ \Longrightarrow 1275\\\ \\

Hence 1275 is the right answer

(03) Find the derivative of following expression:
y=( x-1) \ \left( x^{2} +3\right)

Solution
y=( x-1) \ \left( x^{2} +3\right) \\\ \\ Solving\ the\ equation\\\ \\ y\ =\ x\left( x^{2} +3\right) -1\left( x^{2} +3\right)\\\ \\ y\ =\ x^{3} +3x-x^{2} -3\\\ \\ y\ =\ x^{3} -x^{2} +3x-3 \\\ \\ Differentiating\ the\ equation\\\ \\ \frac{dy}{dx} =3x^{2} -2x+3 \\\ \\

Hence 3x^{2} -2x+3 is the right answer

(04) find the derivative of following expression
y=\left( 2x^{2} -4x\right) .\sqrt{x}\

Solution
Here we will use product of derivative formula

Let\ u\ and\ v\ be\ the\ functions\\\ \\ \frac{d}{dx}[ u.v] =u'v+uv'\\\ \\ For\ this\ question,\ Let:\\\ \\ u\ =\ 2x^{2} -4x\\ \\ u'=4x-4\\\ \\ v=\sqrt{x} \ =( x)^{\frac{1}{2}}\\\ \\ v'=\frac{1}{2}( x)^{\frac{1}{2} -1} \Longrightarrow \ \frac{1}{2.\sqrt{x}}\\\ \\ Putting\ these\ values\ in\ the\ formula\\\ \\ \frac{dy}{dx} =( 4x-4)\sqrt{x} \ +\frac{\left( 2x^{2} -4x\right)}{2\sqrt{x}}\\\ \\ \frac{dy}{dx} \ =\ \frac{( 8x-8) x+2x^{2} -4x}{2\sqrt{x}}\\\ \\ \frac{dy}{dx} \ =\ \frac{8x^{2} -8x+2x^{2} -4x}{2\sqrt{x}}\\\ \\ \frac{dy}{dx} \ =\ \frac{10x^{2} -12x}{2\sqrt{x}}\\\ \\ \frac{dy}{dx} \ =\ ( 5x-6)\sqrt{x} \\\ \\

Hence ( 5x-6)\sqrt{x} is the right answer for this question

(05) Differentiate the following function with respect to x
y\ = \ x^{2} +\ sin\ x\ +\ \frac{1}{x^{2}}

Solution

The\ function\ can\ also\ be\ written\ as:\\\ \\ y=\ x^{2} +\ sin\ x\ +\ x^{-2}\\\ \\ \frac{dy}{dx} =\frac{d}{dx} x^{2} +\frac{d}{dx} sin\ x\ +\frac{d}{dx} x^{-2}\\\ \\ \frac{dy}{dx} =2x\ +\ cos\ x\ -2.x^{-3}\\\ \\ \frac{dy}{dx} =2x+cos\ x\ -\frac{2}{x^{3}}