**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` ) / v^{2}

**Learning polynomial functions:**

`d/dx` (c) = 0

`d/dx` (x) = 1

`d/dx ` (cx) = c

`d/dx` (x^{n}) = nx^{n-1}

`d/dx` (cx^{n}) = ncx^{n-1}

**Learning trigonometric functions:**

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

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

`d/dx ` (tan x) = sec^{2} x

`d/dx` (cot x) = - cosec^{2}x

`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)`

A few example problems are given below for learning derivatives.

**Example 1:**

**Find the derivative of the function y = 8x ^{4} + 2x^{2} + 7.**

**Solution:**

Step 1: Given function

y = 8x^{4} + 2x^{2} + 7

Step 2: Differentiate the given function y = 8x^{4} + 2x^{2} + 7 ^{ } with respect to ' x ', to get `(dy)/dx`

`(dy)/dx` = 32x^{3} + 4x

**Example 2:**

**Find the second order derivative of the function y = 4x ^{6} - 8x^{3} + 3x - 9.**

**Solution:**

Step 1: Given function

y = 4x^{6} - 8x^{3} + 3x - 9

Step 2: Differentiate the given function y = 4x^{6} - 8x^{3} + 3x - 9^{ } with respect to ' x ', to get `(dy)/dx`

** **
`(dy)/dx` = 24x^{5} - 24x^{2} + 3

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

`(d^2y)/dx^2` = 120x^{4} - 48x

**Example 3:**

**Find derivative of the function cos ^{3} x.**

**Solution:**

Step 1: Given

y = cos^{3} x

Step 2: Assign variables

Putting cos x = t, we get

y = t^{3} and t = cos x

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

`dy/dt` = 3t^{2} and `dt/dx` = - sin x

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

`dy/dx` = `dy/dt` * `dt/dx`

=- 3t^{2} sin x

= (- 3 sin x)t^{2}

= (- 3 sin x cos^{2}x)

Step 5: Solution

Therefore, **`dy/dx` = (- 3 sin x cos ^{2}x)**