Introduction to learning derivatives:

In mathematics, the derivative is a measure of how a function changes as its input changes. The process of finding a derivative is called differentiation. A derivative can be thought of as how much one quantity is changing in response to changes in some other quantity. The reverse process is called anti-differentiation.

Some common derivative formulas and example problems are given below which helps you for learning derivatives.                                                                                 (Source: Wikipedia)   

 

 

Learning common derivative formulas:

Learning algebraic functions:

`d/dx` (u + v) = `(du)/dx` +` (dv)/dx`

`d/dx` (u - v) = `(du)/dx` - `(dv)/dx`

`d/dx` (uv) = u `(dv)/dx` + v `(du)/dx`

`d/dx` (u/v) = (v `(du)/dx` - u `(dv)/dx` ) / v2

Learning polynomial functions:

`d/dx` (c) = 0

`d/dx` (x) = 1

`d/dx ` (cx) = c

`d/dx` (xn) = nxn-1

`d/dx` (cxn) = ncxn-1

Learning trigonometric functions:

`d/dx` (sin x) = cos x

`d/dx` (cos x) =  - sin x

`d/dx ` (tan x) = sec2 x

`d/dx` (cot x) = - cosec2x

`d/dx` (sec x) = sec x tan x

`d/dx` (cosec x) = - cosec x cot x

Learning inverse trigonometric functions:

`d/dx` (sin-1 x) = `1/sqrt(1 - x^2)`

`d/dx` (cos-1 x) = - `1/sqrt(1 - x^2)`

`d/dx` (tan-1 x) = `1/(1 + x^2)`

`d/dx` (sec-1 x) = `1/(|x|sqrt(x^2 - 1))`

`d/dx` (cosec-1 x) = - `1/(|x|sqrt(x^2 - 1))`

`d/dx` (cot-1 x) = - `1/(1 + x^2)`

 

Learning derivative with example problems:

A few example problems are given below  for learning derivatives.

Example 1:

Find the derivative of the function y = 8x4 + 2x2 + 7.

Solution:

Step 1: Given function

y = 8x4 + 2x2 + 7

Step 2: Differentiate the given function y = 8x4 + 2x2 + 7   with respect to ' x ', to get `(dy)/dx`

`(dy)/dx` = 32x3 + 4x

Example 2:

Find the second order derivative of the function y = 4x6 - 8x3 + 3x - 9.

Solution:

Step 1: Given function

y = 4x6 - 8x3 + 3x - 9

Step 2: Differentiate the given function y = 4x6 - 8x3  + 3x - 9  with respect to ' x ', to get `(dy)/dx`

                                 `(dy)/dx` = 24x5 - 24x2 + 3

      Step 3: Again differentiate the above function with respect to ' x ',

                                `(d^2y)/dx^2` = 120x4 - 48x

 

Example 3:

Find derivative of the function cos3 x.

Solution:

Step 1: Given

y = cos3 x

Step 2: Assign variables

Putting  cos x = t, we get

y = t3 and t = cos x

Step 3: Differentiate the above function y = t3 and t = cos x with respect to ' t ' and 'x' respectively, we get

`dy/dt` = 3t2 and `dt/dx` = - sin x

Step 4: Derivative `dy/dx` is obtained by multiplying the above two functions

`dy/dx` = `dy/dt`   * `dt/dx`

=- 3t2 sin x

= (- 3 sin x)t2

= (- 3 sin x cos2x)

Step 5: Solution

Therefore, `dy/dx` = (- 3 sin x cos2x)