**Differentiation formulas :**

Here we are going to see list of formulas used in differentiation.

d (x d (log x) d (Constant) d (√x) d (e d (e d (sin x) d (sin ax) d (cos x) d (cos ax) d (tan x) d (tan ax) d (sec x) d (sec ax) d (cot x) d (cot ax) d (cosec x) d (cosec ax) d (sin d (cos d (tan d (cosec d (sec d (cot d (a
d (uv)
d (u/v) |
n x 1/x 0 1/2√x e ae cos x a cos ax -sin x -a sin ax sec a sec sec x tan x a sec ax tan ax -cosec -cosec -cosec x cot x -a cosec ax cot ax 1/√(1-x -1/√(1-x 1/(1+x -1/(x√(x 1/(x√(x -1/(1+x a u v' + v u' (vu' - uv')/v |

Differentiation using first principles :

Formulas in limits :

**Example 1 :**

Differentiate x⁵ tan x

**Solution:**

Let y = x⁵ tan x

u = x⁵ v = tan x

u' = 5x⁴ v' = sec² x

**(UV)' = UV' + VU'**

= (x⁵)sec² x + (tan x)(5x⁴)

= x⁵sec² x + 5x⁴tan x

= x⁴[xsec² x + tan x]

**Example 2 :**

**Differentiate (x² - 1)/ (x² + 1) with respect to x**

**Solution :**

let y = (x² - 1)/ (x² + 1)

u = x² - 1 v = x² + 1

u' = 2x - 0 v' = 2x + 0

u' = 2x v' = 2x

So y' = [(x² + 1) (2x) - (x² - 1)(2x)] /(x² + 1)²

= [(2x)(x² + 1) - (2x)(x² - 1)] /(x² + 1)²

= [(2x³ + 2x) - (2x³ - 2x)] /(x² + 1)²

= [2x³ + 2x - 2x³ + 2x] /(x² + 1)²

= 4x /(x² + 1)²

**Example 3 :**

Differentiate log (sin x) with respect to x

**Solution :**

let y = log (sin x) and we are going to take u = sin X

Now the function becomes y = log u

**dy/dx = (dy/du) x (du/dx)**

dy/du = 1/u

du/dx = cos X

dy/dx = (1/u) x cos X

= cos X/u

= cos X/sin X

= cot X

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